Plane  and  Solid 

Geometry 

Suggestive  Method 

REVISED  EDITION 

By 
George   C.  Shutts 

Instructor  in  Mathematics,  State  Normal  School 
Whitewater,  Wisconsin 


Atkinson,  Mentzer  &  Company 

Boston  New  York  Chicago  Dallas 


Copyright,  1912 
BY  GEORGE  C.  SHUTTS 


TABLE  OP  CONTENTS 

Plane  Geometry: 

Preface  

Introduction    , 1 

Axioms  and  Postulates 11 

Symbols   and   Abbreviations 13 

Rectilinear  Figures — 

Triangles   19 

Quadrilaterals 52 

Polygons  62 

Locus    65 

Circles 73 

Construction  Problems 87,  16? 

Measurement  and  Proportion 97 

Trigonometric  Relations 121 

Measurement  of  Angles 126 

Extreme  and  Mean  Ratio 145 

Numerical  Relations 147 

Area  of  Polygons 1  fi ") 

Regular  Polygons 191 

Incommensurable  Magnitudes   21 3 

Solid  Geometry: 

Lines  and  Planes 223 

Dihedral  Angles 248 

Polyhedral  Angles 260 

Polyhedrons   270 

Prisms    271 

Pyramids 290 

Similar  Polyhedrons   302 

Regular   Polyhedrons    306 

Cylinders 309 

Cones    I 318 

Spheres    330 

Index  .  . .375 


PREFACE. 

In  this  revision  the  suggestive  features  of  former 
editions  have  been  retained,  and  it  is  hoped  improved. 
The  principal  value  in  the  study  of  geometry  lies  in 
the  power  developed  by  the  individual  student  in  work- 
ing out  as  much  as  possible  his  own  demonstrations. 

Pupils  in  their  early  study  of  the  subject  of  geom- 
etry sometimes  fail  to  see  the  need  of  some  of  the  steps 
in  a  rigorous  deductive  demonstration  and,  to  satisfy 
the  demands  of  the  class  room,  resort  to  formal  memory 
of  text  as  a  substitute  for  logical  thinking.  This  text 
has  been  prepared,  not  so  much  to  illustrate  a  logical 
development  of  mathematical  science  as  to  arrange  and 
adapt  the  mathematical  data  herein  contained  to  the 
comprehension  and  growth  of  the  pupil.  In  the  intro"- 
duction  of  various  subjects  theorems  are  stated  as  postu- 
lates or  preliminary  propositions  which  in  a  strictly 
logical  development  would  require  proof.  Many  of  these 
statements  are  of  so  fundamental  a  nature,  have  so 
long  been  accepted  as  obvious  facts  by  the  pupil,  and 
yet  are  so  difficult  of  demonstration,  that  the  chief 
value  in  the  use  of  them  lies  not  so  much  in  their  demon- 
stration as  in  the  comprehension  and  use  of  them  in 
the  demonstration  of  truths  based  upon  them.  Theo- 
rems have  been  chosen  for  the  earlier  demonstrations 
such  that  the  course  of  reasoning  will  appeal  to  the 
logical  ability  of  the  pupils  at  that  stage  of  their  rea- 
soning power.  ^59793 


The  protractor  T  square,  parallel  rulers/  etc.,  have 
been  brought  into  use  before  the  principles  upon  which 
their  construction  is  based  have  been  developed,  that 
through  a  mechanical  construction  of  the  various  forms, 
the  pupil  may  get  a  better  understanding  of  their  mean- 
ing than  he  usually  gets  from  abstract  definitions. 

The  rigor  of  the  demonstrations  has  in  no  case  been 
minimized  and  constructions,  when  the  principles  upon 
which  they  depend  have  been  developed,  as  usual  require 
the  use  of  straight  edge  and  dividers  only. 

The  subject  of  measurement  and  proportion  has 
been  simplified,  and  the  incommensurable  cases  placed 
in  the  latter  part  of  the  text  in  plane  geometry,  with 
the  intent  that  in  a  short  course  in  plane  geometry  this 
phase  may  be  omitted  if  the  teacher  deems  it  wise  to 
do  so. 

Many  practical  applications  of  the  principles  of 
geometry  have  been  introduced  in  the  exercises,  but 
have  been  selected  with  reference  to  the  interests  of  all 
the  pupils,  rather  than  the  few  who  desire  to  enter 
technical  occupations.  Many  of  the  old  geometrical  puz- 
zles have  been  omitted  and  numerical  problems  of  in- 
terest introduced.  It  is  not  expected  that  any  given 
class  or  pupil  will  work  all  of  the  exercises,  but  the 
number  is  large  enough  and  the  variety  sufficiently  great 
to  furnish  interesting  recreation  and  drill  for  the  var- 
iety of  tastes  in  any  class  of  students.  Many  exercises 
follow  the  propositions  to  which  they  are  related  for 
application  and  drill,  but  many  others  are  scattered 
throughout  the  text  more  or  less  unrelated  to  serve  the1 
purpose  of  review  and  test  the  pupil's  power  of  indepen- 
dent thought. 

The  one  suggestion  the  author  would  make  to  the 
teacher  is  that  success  depends  upon  letlinir  tin1  nilc  <>!' 


progress  be  determined  by  the  thorough  comprehension 
by  the  pupils  of  the  subject  matter.  The  amount  of 
text  covered  is  not  so  important  as  the  real  develop- 
ment of  the  pupils. 

The  author  wishes  to  acknowledge  for  their  scholarly 
assistance,  his  obligation  to  H.  D.  Merrill  of  the  chair  of 
mathematics  of  the  Evanston  High  School,  who  has 
made  valuable  suggestion  throughout ;  to  A.  W.  Smith, 
Professor  of  Mathematics  of  Colgate  University,  Ham- 
ilton, N.  Y.,  who  assisted  in  the  revision  and  read  the 
proof,  and  to  many  students  who  through  conscientious 
work  in  the  class  room  have  given  valuable  suggestions. 
To  these  old  friends  and  to  the  prospective  students,  to 
whom  it  is  hoped  these  pages  will  furnish  inspiration, 
this  work  is  dedicated. 

GEORGE  C.  SHUTTS. 

Whitewater,  Wis.,  Aug.  1,  1912. 


PLANE  GEOMETRY 


INTRODUCTION. 

1.  PHYSICAL  SOLID.     Any  material  object  is  a  phys- 
ical solid,  for  example,  a  block  of  wood  or  a  ball. 

2.  GEOMETRIC  SOLID.     A  limited  portion  of  space  is 
a  geometric  solid.     A  block  of  wood  is  a  physical  solid 
but  the  portion  of  space  occupied  by  it  is  a  geometric* 
solid.     Two  reasons  are  suggested  for  this  definition  of 
a  geometric  solid:  first,  geometry  is  not  concerned  with 
the  material  of  which  an  object  is  made  but  with  its 
size  and  its  shape ;  second,  such  solids  can  be  made  to 
coincide  in  whole  or  in  part. 

3.  SURFACE.     That  which  has  length   and  breadth 
without  thickness  is  a  surface. 

The  boundaries  cf  a  solid  are  surfaces  but  these  surfaces  are 
not  a  part  of  the  solid  which  they  bound  any  more  than  one's 
shadow  on  a  wall  is  a  part  of  the  wall.  A  surface  may  pass 
through  a  solid  or  another  surface  and  may  be  unlimited  in 
extent. 

4.  LINE.     That  which  has  length  without  breadth  or 
thickness  is  a  line. 

A  line  may  pass  through  a  solid,  a  surface,  or  another  line  and 
may  be  unlimited  in  extent. 

5.  POINT.     That  which  has  position  without  dimen- 
sions is  a  point. 


2  PLANE  GEOMETRY 

6.  MAGNITUDES.     Lines,    surfaces,    and    solids    are 
geometric  magnitudes. 

7.  MAGNITUDES  GENERATED.     The  path  of  a  moving 
point  is  a  line,  or  a  point  in  motion  generates  a  line.     A 
line  in  motion  (except  upon  itself)  generates  a  surface. 
A  surface  in  motion   (except  upon  itself)   generates  a 
solid. 

8.  MAGNITUDES,    LIMITED   OR    BOUNDED.     Solids   are 
limited  or  bounded  by  surfaces,  surfaces  by  lines,  lines 
by  points,  unless  in  accordance  with  the  definitions  given 
above  they  are  unlimited. 

9.  GEOMETRIC  FIGURES.     Any  combination  of  points, 
lines,  surfaces,  and  solids  is  a  geometric  figure.      Some 
Geometric  figures  are  mental  images  only.    They  cannot 
be  constructed  and  the  drawings  used  are  but  represen- 
tations.    For  example,  a  point  may  be  represented  by 
a  crayon  or  pencil  dot,  a  line  by  a  pencil  mark  or  the 
edge  of  a  ruler,  a  surface  by  a  table  top  or  the  surface 
of  a  blackboard,  etc. 

10.  STRAIGHT  LINE.     A  line  is  straight  if  any  part 
of  it  will  lie  wholly  in  any  other  part  when  its  extrem- 
ities lie  in  that  part,  or  if,  when  revolved  with  two  points 
kept  stationary,  it  occupies  the  same  position  as  before. 
A  straight  line  may  be  thought  of  as   extending  in- 
definitely in  both  directions.     The  first  test  may  be  ap- 
plied to  a  pencil  mark  representing  a  straight  line  by 
using  tracing  paper  to  apply  "any  part"  to  "any  other 
part"  of  it  or  by  putting  together  the  edges  of  two 
rulers  as  representing  two  parts  of  a  straight  line.     The 
second   test  may  well  be   illustrated   with   a   piece  of 
straight  wire. 


INTRODUCTION  3 

11.  LINE  SEGMENT.     A  limited  portion  of  a  line  is  a 
line-segment. 

A  line-segment  is  sometimes  called  a  sect. 

12.  RAY.     A  portion  of  line  limited  at  one  end  only 
is  a  ray.    The  limiting  point  of  a  ray  is  its  origin.     A 
segment  differs  from  a  ray  in  that  it  has  two  limiting 
points  instead  of  one. 

13.  NOTATION.     A  point  is  read  by  naming  a  capital 
letter  placed  near  it.     A  line  is  read  by  naming  the  let- 
ters which  mark  any  two  of  its  points.     Lines,  rays,  and 
line-segments  are   at   times   indicated   by   single   small 
letters  instead  of  two  capitals  indicating  points.     For 
example  in  the  following  figures  one  reads :  point  A,  line 
BC  or  I,  segment  DE  or  s,  ray  FG  or  r. 


14.  EQUAL  SEGMENTS.     If  two  line  segments  can  be 
so  placed  that  their  extremities  coincide  the  segments 
are  said  to  be  equal. 

15.  ADDITION  AND  SUBTRACTION  OF  SEGMENTS. 

(1)  Two  segments  are  added  by  placing  them  end 
to  end.     The  resulting  segment  is  called  the  sum. 

(2)  Two   segments   are   subtracted  by   placing  the 
shorter  upon  the  longer  so  that  one  pair  of  end  points 
coincide.     That  portion  of  the  longer  not  covered  by  the 
shorter  is  called  the  difference. 

For     example,          A BC D 

to  find  the  sum          A^  D      B- v 

of    segment    AB  ~  ~~* 

and  segment  CD,  draw  an  indefinite  line  and  on  it  lay  off 
A'B'  equal  to  AB  and  B'D'  equal  to  CD.  Then  A'D'  is 
the  sum  of  AB  and  CD  or  as  it  may  be  written  in  symbols, 


•±  PLANE  GEOMETRY 

AB  +  CD  =--  A'D'.  To  find  the  difference  between  AB 
and  CD  lay  CD  upon  AB  with  point  C  on  A.  The  dif- 
ference is  the  segment  DB.  The  relation  is  here  written 
as  AB  —  CD  =  DB. 

The  measurements  here  necessary  may  be  made  with  a 
ruler  or  scale  or  better  yet  with  a  pair  of  dividers. 

16.  BROKEN  LINE.    A  line  made 
up  of  successive  segments  not  form- 
ing a  straight  line  is  a  broken  line. 

17.  CURVED    LINE.     A    line    no    part    of    which    is 
straight  is  a  curved  line  or  simply  a  curve. 

18.  PLANE.     A    surface    such    that    a    straight    line 
through  any  two  of  its  points  lies  wholly  in  the  surface 
is  a  plane.     A  plane  is  sometimes  spoken  of  as  a  flat  sur- 
face and  portions  of  a  plane  may  be  enclosed  by  broken 
lines  or  curves. 

19.  PLANE  FIGURES.     A  figure  all  parts  of  which  lie 
in  the  same  plane  is  a  plane  figure. 

(1)  A  figure  all  lines  of  which  are  straight  is  a 
rectilinear  figure. 

(2)  A  portion  of  a  plane   entirely   enclosed  by  a 
broken  line  is  a  plane  polygon  or  simply  a  polygon. 

(3)  A  curve  enclosing  a  portion  of  a  plane  in  such  a 
manner  that  every  point  of  the  curve  is  at  the  same  dis- 
tance   from    a    point    within    is    a    circle,    the    point 
within  is  the  center,  the  distance  from  the  center  to  the 
circle  is  the  radius,  and  any  portion  of  the  circle  is  an  arc. 

20.  PLANE     GEOMETRY.     That     part     of     geometry 
which  treats  of  plane  figures  is  called  plane  geometry. 

21.  ANGLE.     Two   rays   proceeding    from   the    same 
point  contain  or  make  an  angle.    The  rays  are  the  sides 


INTRODUCTION  5 

of  the  angle  and  the  point  of  meeting  is  the  vertex  of 
the  angle. 

An  angle  may  be  read  by  naming  the  letters  which 
designate  its  sides,  the  vertex  letter  being  read  between 
the  others,  as  the  angle  AOB.  If  no 
confusion  results,  an  angle  may  be 

read  by  naming  its  vertex  letter  alone,      Q^—J^. A 

as  the  angle  0,  or  by  naming  a  single  small  letter,  as 
angle  m. 

22.  TURNING  A  LINE  THROUGH  AN  ANGLE.     If  a  line 
coincident  with  one  side  of  an  angle  be  revolved  about 
the  vertex  as  a  pivot  until  it  coincides  with  the  other  side 
of  the  angle,  the  line  turns  through  the  angle.     This 
turning  is  usually  conceived  to  be  counter-clockwise. 

23.  SIZE  OP   ANGLE.     The   size   of  an   angle   is  the 
amount  of  revolution  necessary  to  turn  a  line  through  it. 
It  bears  no  relation  to  the  length  of  the  sides. 

24.  EQUAL  ANGLES.     If  two  angles  can  be  placed  so 
that  their  vertices  coincide  and  the  sides  of  the  one  lie 
upon  the  sides  of  the  other  the  angles  coincide  and  the 
turning  through  both  angles  is  effected  at  the  same  time. 
Such  angles  are  said  to  be  equal. 

A  pair  of  dividers  may  be  used  to  represent  an  angle,  the  size 
of  the  angle  depending  upon  the  extent  to  which  the  dividers  are 
opened.  They  may  be  used  to  compare  two  angles  by  placing 
them  upon  one  of  the  angles  with  the  legs  of  the  dividers  lying 
along  the  sides  of  the  angle  and  then  placing  them  upon  the  other 
angle,  noticing  whether  or  not  it  is  necessary  to  open  or  close 
them  in  order  that  they  fit  the  second  angle.  A  similar  compari- 
son may  be  made  by  drawing  one  angle  on  a  piece  of  tracing 
paper  and  placing  it  upon  the  other  angle  so  that  the  vertex  and 
one  side  of  the  trace  lie  upon  the  vertex  and  one  side  of  the 
second  angle.  The  relative  position  of  the  remaining  sides  of 
the  two  angles  show  which  is  the  larger. 


PLANE  GEOMETRY 


25.  ADJACENT  ANGLES.     Two  an- 
gles that  have  the  same  vertex  and 
one   common  side  between  them  are 
adjacent  angles. 

The  angles  AOB  and  BOC  are  adjacent. 

26.  ADDITION  AND  SUBTRACTION  OF  ANGLES. 

(1)  Two  angles  are  added  by  so  placing  them  as  to 
form  adjacent  angles.     The  sum  is  the  large  angle  thus 
formed. 

(2)  Two  angles  are  subtracted  by 
placing  the  smaller  within  the  larger 
so  that  they  have  one  side  and  the 
vertex  in  common.     The  angle  adja- 
cent to  the  smaller  is  the  difference. 
For  example,  the  sum  of  angles  AOB 
and  BOC  is  AOC  and  the  difference 
between  AOB  and  DOB  is  AOD. 

Are  these  results  consistent  with 
the  statement  in  §  22?  By  means  of 
tracing  paper  obtain  the  sum  of  the 
angles  AOB  of  §  21  and  BOC  of  §  25. 

27.  PERIGON.     The  total  angular  magnitude  about  a 
point  in  a  plane  is  a  perigon.    A  line  has  turned  through 

a  perigon  when  it  has  made  a  complete  ,••"•""••.  _ 

*  o     ^ 
revolution  about  a  point  as  BOB'.  {       7fT~ 

28.  STRAIGHT  ANGLE.     An  angle  the  sides  of  which 
lie  in  a  straight  line  and  on  opposite  sides  of  the  vertex 
is  a  straight  angle,  as  angle  AOB.  /,*--% 

A  straight  angle  is  one  half  a  peri-      B  o          "~A 

gon.    The  student  must  distinguish  between  the  straight 
line  AOB  and  the  straight  angle  A  OB. 


INTRODUCTION  7 

Let  the  student  turn  a  ray  through  the  angle  AOB. 

29.     RIGHT  ANGLE.     If  a  ray  meets  a  line  so  that  the 
two   angles   formed   are   equal,    each 
angle  is  a  right  angle,  as  L  AOB  and 
LEOC. 


o 

(1)  A  right  angle  is  one  half  of  a  straight  angle. 

(2)  A  right  angle  is  one  fourth  of  a  perigon. 

30.  DEGREE.  One  ninetieth  part  of  a  right  angle  is 
a  degree. 

(1)  One  sixtieth  part  of  a  degree  is  a  minute  and 
one  sixtieth  part  of  a  minute  is  a  second. 

(2)  Degrees,  minutes,  and  seconds  are  denoted  by 
the  symbols  °,  ',  "  respectively,  as  14°  27'  30". 

(3)  The   degree   is   a   unit  of  measure   for  angles. 
A   protractor   is   sometimes   used   in   constructing   and 
measuring  angles.     It  consists  of  a  semicircular  scale 
with  degrees  from  0°  to  180°  marked  on  it.     To  use  the 
protractor  in  measuring  an  angle  place  the  center  0  at 
the  vertex  of  the  angle  and  the  zero  of  the  protractor 
upon  one  side  of  the  angle.    The  number  of  degrees  and 
minutes  in  the  angle  is  the  number  indicated  upon  the 
protractor  at  the  point  where  the  second  side  of  the  angle 
projects  beyond  the  protractor.    The  protractor  can  also 
be  used  in  constructing  an  angle  of  a  given  number  of 
degrees  or  one  that  is  equal  to  a  given  angle. 

With  the  protractor  construct  a  right  angle,  cut  it  out,  and  test 
the  accuracy  of  the  construction  by  superimposing  it  upon  a 
known  right  angle. 


8  PLANE  GEOMETRY 

31.  ACUTE  ANGLE.     An  angle  less  than 
a  right  angle  is  an  acute  angle. 

With    a    protractor    construct    acute    angles    of 
20°,  70°,  65°,  40°,  28°,  53',  etc. 

32.  OBTUSE     ANGLE.     An     angle 
greater  than  a  right  angle  and  less 
than    a  straight   angle   is   an  obtuse 
angle. 

With  a  protractor  construct  obtuse  angles  of  95°,  115°,  170°. 
Add  angles  of  82°  and  32°;  of  47°  and  53°. 

33.  REFLEX    ANGLE.      An    angle 

greater  than  a  straight  angle  and  less  i^ — . 

than  a  perigon  is  a  reflex  angle.  ^^ 

Reflex  angles  are  seldom  considered  in  elementary  geometry. 

With  a  protractor  or  tracing  paper  add  angles  of  98°  and  71° ; 
of  120°  and  92°;  of  72°  and  65°.  What  kind  of  angles  are  the 
sums? 

34.  OBLIQUE  ANGLES.     Acute  and  obtuse  angles  are 
oblique  angles. 

35.  COMPLEMENTS.     Two  angles  the  sum  of  which  is 
a  right  angle  are  complements. 

With  a  protractor  construct  the  complements  of  60°,  50°,  75°, 
49°.  Measure  each  and  test  the  results. 

36.  SUPPLEMENTS.     Two  angles  the  sum  of  which  is 
a  straight  angle  are  supplements.     The  complement  or 
supplement  of  a  given  angle  need  not  be  adjacent  to  it. 

Construct  the  supplement  of  18°,  24°,  96°,  125°.  Which  are 
acute  and  which  obtuse?  Test  the  accuracy  of  the  results  with 
the  protractor. 

37.  CONJUGATES.     Two  angles  the  sum  of  which  is  a 
perigon  are  conjugates. 

Two  rays  proceeding  from  a  point 
really  form  two  angles,  as  the  angles  ^ 

m  and  n.     These  angles  are  conju-    /*"*N-^ 
gates. 


INTRODUCTION  0 

Measure  the  degrees  in  m  and  //  and  to^t  the  result  by  this 
definition. 

In  referring  to  an  angte  the  smaller  of  the  two  conju- 
gates is  meant  unless  otherwise  stated. 

The  student  should  provide  himself  with  a  ruler,  a  protractor, 
a  pair  of  dividers,  parallel  rulers,  and  if  convenient  a  T  square 
and  drawing  board. 

1.  In  the  adjacent  figure  AE  is  a 
straight  line  and  CO  is  perpendicular 
to   AE.      Read   a   straight  angle,   two 
right  angles,  five  acute  angles,  two  ob- 
tuse angles.  o~ 

2.  Eead  ten  pairs  of  adjacent  angles  in  the  above  figure. 

3.  What  is  the  sum  of  angles  x  and  u?     Of  angles  y  and  v? 
Of  angles  u  and  x?     Of  angles  y  and  x? 

4.  Read    two    pairs    of    complementary    angles    in    the    above 
figure. 

5.  Read   three   pairs   of   supplementary    angles   in   the   above 
figure. 

6.  A    perigou    is    equal    to    how    many    straight    angles?      A 
straight  angle  is  equal  to  how  many  right  angles!     A  perigon  is 
equal  to  how  many  right  angles?     Compare  a  straight  angle  with 
a  perigon ;   a  right  angle  with  a  perigon ;    a  right   angle  with  a 
straight  angle. 

7.  How  many  degrees  are  there   in  a   straight  angle?     In  a 
perigon? 

8.  How  many  degrees  in  the  supplement  of  a  right  angle?     In 
the  supplement   of  two-thirds  of  a  right  angle?     In  the  comple- 
ment of  three-fourths  of  a  right  angle? 

9.  Find  the  supplement  and  the  complement  of  54°   27' ;   of 
57°  35'  42". 

10.  How   many   degrees   in    tha  smaller   angle   formed   by   the 
hands  of  a  clock  at  1  o'clock?     At  5  o'clock?     How  many  degrees 
are  generated  by  the  minute  hand  of  a  clock  from  two  o  'clock   to 
twenty  minutes  past  two? 

11.  A   perigon    is   divided   into   six   equal  angles.      How   many 
degrees  are  there  in  each?     Illustrate  with  a   figure. 


10  PLANE  GEOMETRY 

12.  A  perigon  is  divided  into  three   angles,  the  second   being 
twice  the  first  arid  the  third  three  times  the  first.     Find  the  num- 
ber of  degrees  in  each  angle.     Illustrate  with  a  figure. 

Suggestion:  Let  x  equal  the  number  of  degrees  in  the  first. 

13.  Write  an  equality  that  will  say  that  angles  x  and  y  are 
complements;  that  they  are  supplements. 

14.  Angles  x   and   y   are   supplements   and   their   difference   is 
50°.     Find  angle  x  and  angle  y.     With  a  protractor  draw  x  and  y 
and  check  the  results. 

15.  How  can  the  supplement  of  an  angle  be  found?     Draw  an 
angle,  as  x,  with  the  protractor  and  find  its  supplement. 

16.  The  supplement  of  an  angle  x  is  three  times  the  comple- 
ment of  x.     Find  the  angle  x. 

17.  In  the  figure  of  example  1  measure  with  a  protractor  the 
angles  x,  y,  and  z ;  add  them  and  test  the  result  by  measuring 
angle  ADD. 

18.  Draw    a    straight    angle.      Divide   it    by    a    ray    into    two 
oblique  angles.     Measure  each  and  add  them.     How  many  degrees 
should  their  sum  be? 

19.  Divide  a  perigon  into  five  angles  and  measure  each.     Test 
the  accuracy  of  the  work  by   adding   the  re- 
sults. 

20.  Measure   angle   A    and   angle   CAD. 
Subtract   the   results   and   test  the   accuracy 
of  the  work  by  measuring  angle  DAB. 

21.  Draw    a    line-segment    1    inch    long.      Draw    another    1$ 
inches  long  so  as  to  include  between  them  an  angle  of  45°.     Then 
measure   the    distance    between    the    free    ends    of    the    segments. 
Make  the  construction  a  second  time  and  with  tracing  paper  test 
whether  the  two  figures  are  congruent.     §  64. 

22.  Draw  an  angle   of  25°,  18°,  72°,   126°,   175°. 

23.  Draw  a  reflex  angle  of  197°,  240°,  315°,  345°. 

38.     VERTICAL  ANGLES.    When  the 
sides  of  one  angle  are  extensions  of  c 
the  sides  of  another  angle,  the  two 
angles  are  vertical  angles. 

For  example,  m  and  n  are  vertical  angles. 


INTRODUCTION  11 

39.  PERPENDICULAR  LINES.     If  two  lines  form  a  right 
angle,  each  line  is  perpendicular  to  the  other.    The  word 
perpendicular  is  used  also  as  a  noun. 

40.  OBLIQUE  LINES.     If  two  lines  form  an  oblique 
angle,  each  line  is  oblique  to  the  other. 

41.  BISECTOR.     If  a  figure  is  divided  into  two  equal 
parts  it  is  bisected. 

In  plane  geometry  the  bisector  of  a  segment  may  be  a  point, 
a  segment,  a  ray,  or  a  line.  The  bisector  of  an  angle  may  be  a 
segment,  a  ray,  or  a  line. 

42.  PROOF.     To  prove  a  statement  is  to  show  by  a 
logical  course  of  reasoning  that  it  must  be  true  by  means 
of  other  statements  that  have  been  accepted.    In  general 
the  proof  of  a  statement  involves  other  statements  which 
in  turn  require  proof.     Evidently  as  a  foundation  for 
reasonings  some  statements  must  be   accepted  without 
proof. 

43.  AXIOM.     A  statement  admitted  without  proof  to 
be  true  and  not  limited  to  geometric  figures  is  an  axiom. 

44.  POSTULATE.     A  statement  admitted  without  proof 
and  limited  to  geometric  figures  is  a  postulate. 

45.  THEOREM.     A  statement  that  is  to  be  proved  is 
a  theorem. 

46.  PROBLEM.     The   statement   of   a   geometric   con- 
struction to  be  made  is  a  problem. 

47.  PROPOSITION.     Theorems  and  problems  are  prop- 
ositions. 

48.  COROLLARY.     A  proposition  easily  deduced  from 
another  proposition  is  a  corollary. 

49.  AXIOMS.     In  the  following  axioms  the  equalities 
and  inequalities  concerned  are  unconditional. 

(1)  Quantities  that  are  equal  to  the  same  quantity, 
or  to  equal  quantities,  are  equal  to  each  other. 


12  PLANE  GEOMETRY 

(2)  If  equals  are  added  to  or  subtracted  from  equals 
the  results  are  equal. 

(3)  If  equals  are  multiplied  or  divided  by  equals 
(division  by  zero  being  excluded)  the  results  are  equal. 

(4)  If  equals  are  added  to  or  subtracted  from  un- 
equals  the  results  are  unequal  in  the  same  order. 

(5)  If  unequals  are  multiplied  or  divided  by  posi- 
tive equals  the  results  are  unequal  in  the  same  order. 

(6)  If  unequals  are  added  to  unequals  in  the  same 
order  the  results  are  unequal  in  that  order. 

(7)  If  unequals  are  subtracted  from  equals  the  re- 
sults are  unequal  in  the  reverse  order. 

(8)  If  the  first  of  three  quantities  is  greater  than 
the  second  and  the  second  is  greater  than  the  third,  then 
the  first  is  greater  than  the  third. 

(9)  In  considerations  involving  size  only,  the  whole 
is  greater  than  any  of  its  parts  and  is  always  equal  to 
the  sum  of  its  parts. 

(10)  Either  of  two  equals  may  be  substituted  for  the 
other  in  any  process. 

(11)  Every     magnitude,     however     small,     can     be 
divided  into  two  or  more  parts. 

50.     POSTULATES. 

(1)  If  two  straight  lines  have  two  points  in  com- 
mon they  become  one  and  the  same  straight  line. 

Between  two  points  there  is  one  and  only  one  straight  line. 

(2)  The   shortest   path   between   two  points   is  the 
straight  line  segment  joining  them. 

(3)  A  line-segment  has  one  and  but  one  mid-point. 
An  angle  has  one  and  only  one  bisector. 

(4)  A  figure  can  be  moved  without  altering  its  size 
or  its  shape. 

(5)  A  line  in  a  plane  divides  the  points  of  that  plane 


INTRODUCTION 


13 


into  two  classes  such  that  a  line-segment  connecting  two 
points  of  the  same  class  is  not  intersected  by  the  line  • 
and   a   line-segment   connecting  two   points   not  of   the 
same  group  is  intersected  by  the  line. 

(6)  Magnitudes  which  coincide  are  equal. 

(7)  All  perigons  are  equal. 

(8)  A  circle  or  an  arc  can  be  constructed  with  any 
radius  and  any  given  point  as  center. 

(9)  A  straight  line  can  be  drawn  between  two  points ; 
a  straight  line  can  be  terminated  at  any  point ;  a  sect 
can  be  extended  any  distance. 

SYMBOLS  AND  ABBREVIATIONS. 


The  symbols   +,   — ,    X 
>  is  greater  than. 
<  is  less  than. 
1  perpendicular. 
I!  parallel. 
=  is  equal  to. 
'-'  is  similar  to. 
~  is  congruent  to. 
m  is  measured  by. 
=  approaches  as  a  limit. 
L  angle. 
A  triangle. 
O  parallelogram. 
Adj.  adjacent. 
Alt.  alternate. 
Auth.  authority. 
Ax.  axiom. 
Const,   construction. 
Cor.  corollary. 


^~  are  used  as  in  algebra. 
Def.  definition. 
Ex.  exercise. 
Ext.  exterior. 
Fig.  figure. 
Hyp.  hypothesis. 
O  rectangle. 
O  circle. 
§  article, 
rt.  right, 
st.  straight. 
•"•  therefore. 
*•*  hence. 
Iden.  identity: 
Int.  interior. 
Post,  postulate. 
Prop,  proposition. 
Sug.  suggestion. 
Sup.  supplementary. 


14  PLANE  GEOMETRY 

The  symbols  Z,  A,  O,  CD,  O  have  the  plural  forms 
A,  A,  O7,  m,  ©. 

PRELIMINARY  THEOREMS.  The  following  are  some  im- 
portant preliminary  theorems : 

51.  THEOREM.     1.     Straight  angles  are  equal. 

SUG.     §  28,  Post.  7,  and  Ax.  3. 

52.  THEOREM  2.     Right  angles  are  equal. 

SUG.     §  29  and  Ax.  3. 

53.  THEOREM  3.     (a)     Complements  of  equal  angles 
are  equal,     (b)   Supplements  of  equal  angles  are  equal. 

a.  Given  angles  A  and  B,  each 
being  a  complement  of   Z  C.     To 
prove  Z  A  and  Z  5  equal. 

SUG.  1.  How  does  the  sumA 
of  A  A  and  C  compare  with 
the  sum  of  A.  B  and  (7?    Th.  2  and  Ax.  1. 

2.     Compare  Z  A  and  Z  B.    Ax.  2. 
Therefore — 

b.  Follow  the  method  of  demonstration  used  for  theo- 
rem 3  (a). 

N0te. — Before    looking    up    references    in    the    suggestions    the 
pupil   should   endeavor  to   determine   the   authorities   for   himself. 

54.  THEOREM  4.     Two  right  angles  are  supplements 
of  each  other. 

55.  THEOREM  5.     Only  one  line  can 
be  drawn  perpendicular  to  a  line  at  a 
given  point  on  the  line. 

SUG.     Can  both  OD  and  OC  be 
1  to  AB  at  point  0?     §29,   §39  * 
and  §  50  (3). 

56.  THEOREM  6.     Two  lines  that  intersect  can  have 
but  one  point  in  common. 


INTRODUCTION  15 

SUG.     If  they  had  two  points  in  common,  what 
would  be  true  ?    Post.  1. 

57.     THEOREM  7.     //  two  straight  lines  intersect  the 
vertical  angles  are  equal. 

Given  AB  and  CD  intersect- 
ing at  0  with  vertical  angles  in 
and  p. 
To  Prove  Z  m  =  Z  p. 

SUG.  1.     What  relation  does  Z  m  bear  to  Z  n? 
§36. 

2.  What  relation  does  L  p  bear  to  Z  ?i  ? 
.       §36. 

3.  Compare  Z  m  and  L  p.    §  53  (7J. 
Therefore— 

Compare  angles  q  and  w. 

1.  Draw  two  supplementary  adjacent  angles  and  bisect  each. 
How  many  degrees  in  the  angle   of  the  bisectors?     Use  the  pro- 
tractor and  straight  edge. 

2.  How  many  degrees  in  the  angle  formed  by  the  bisectors 
of    two    complementary    adjacent    angles?      By    the    bisectors    of 
two  adjacent   angles   of  63°   ari  70°  respectively?     Of  26°    and 
84°  respectively? 

3.  If  two  lines  intersect  and  one  of  the  angles  is  a  right  angle, 
prove  the  others  are  right  angles. 

4.  If  one  angle  formed  by  two  intersecting  lines  is  70°   how 
many  degrees  are  there  in  each  of  the  others?     If  one  angle  is 
50°  how  many  in  each  of  the  others? 

5.  The  difference  between  two  complementary  angles  is  12°. 
What  are  the  angles?     If  the  angles  are  supplementary,  what  are 
they? 

6.  Solve  problem  5  if  the  difference  is  40° ;  75° ;  82°  16'. 

7.  The  ratio  of  two  complementary  angles  is  4:5.    What  are 
the  angles?    Solve  for  ratios  of  3:6,  3^:5^,  5:9. 

8.  Solve  problem  7  with  the  supposition  that  the  angles  are 
supplementary. 


16  PLANE  GEOMETRY 

9.  What  is  the  complement  of  100°?  The  supplement  of 
212°?  Of  198°  27'?  If  the  student  by  use  of  algebra  derives 
these  results  from  the  definitions  of  complements  and  supple- 
ments, he  will  have  angles  with  the  negative  sign.  The  interpre- 
tation is  this:  Angles  which  are  generated  by  counterclockwise 
notation  are  called  positive  and  those  generated  by  a  clockwise 
notation  are  called  negative.  This  plan  is  of  great  use  whenever 
it  is  necessary  to  distinguish  between  these  two  ways  of  generat- 
ing an  angle. 

58.     SUMMARY. 

1.  Two  angles  are  equal 

(1)  if  they  can  be  made  to  coincide; 

(2)  if  they  are  equal  to  the  same  angle  or  to 
equal  angles; 

,  (3)  if  they  can  be  obtained  by  adding  equal  an- 
gles to  equal  angles  or  by  subtracting  equal 
angles  from  equal  angles ; 

(4)  if  they  are  doubles  or  halves  of  the  same 
angle  or  of  equal  angles; 

(5)  if  they  are  straight  angles,  right  angles,  or 
vertical  angles; 

(6)  if  they  are  complements  or  supplements  of 
the  same  or  of  equal  angles. 

2.  Two  angles  are  complements  if  their  sum  is  a 
right  angle  or  90°. 

3.  Two  angles  are  supplements  if  their  sum  is  n 
straight   angle,  two  right   angles,   or   180°. 

4.  An  angle  is  a  right  angle 

(1)  if  it  is  one  of  two  equal  adjacent  angles 
formed  by  two  straight  lines; 

(2)  if  it  is  one  half  of  a  straight  angle  or  con- 
tains 90° ; 

(3)  if  its  sides  are  perpendicular  to  each  other; 


INTRODUCTION  17 

(4)     if  it  is  one  of  two  equal  supplementary  an- 
gles. 

5.  An  angle  is  a  straight  angle 

( 1 )  if  its  sides  form  a  straight  line  ; 

(2)  if  it  is  the  sum  of  two  right  angles  or  180° ; 

(3)  if  it  is  one-half  of  a  perigon. 

6.  Two  lines  are  perpendicular  to  each  other  if  they 
form  a  right  angle. 

7.  Two  lines  form  one  straight  line 

(1)  if  they  have  two  points  in  common; 

(2)  if  they  are  the  sides  of  a  straight  angle; 

(3)  if  they  are  perpendicular  to  the  same  line 
at  the  same  point. 

8.  Line  segments  or  sects  can  be  proved  equal  by 
means  of  axioms  1,  2,  3  or  by  showing  that  they 
can  be  so  placed  that  their  end  points  coincide. 

9.  Angles  or  line  segments  can  be  proved  unequal 

by  means  of  axioms  4-9  inclusive  or  postulate  3. 

.    '      -          .'••<. 

>Y>  i  i  2>  -i 


CHAPTER  I. 
RECTILINEAR  FIGURES. 

59.  TRIANGLE.     A  portion  of  a  plane  bounded  by 
three  sects  is  a  triangle.    The  line-segments  are  the  sides 
of  the  triangle,  their  intersections  are  the  vertices,  their 
sum  is  the  perimeter,  the  angles  formed  by  the  sides  are 
the  angles  of  the  triangle,  and  the  three  sides  and  three 
angles  are  the  parts  of  the  triangle. 

60.  TRIANGLES  CLASSIFIED  AS  TO  SIDES.    A  triangle  no 
two  of  the  sides  of  which  are  equal  is  a  scalene  triangle. 
One  with  two  equal  sides  is  an  isosceles  triangle.     One 
with  three  equal  sides  is  an  equilateral  triangle.     In 
the  accompanying  figures  A  ABC  is  scalene,  A  DEF  is 
isosceles,  and  A  GHI  is  equilateral. 

Measure  to  verify. 

C  F  I 


61.  TRIANGLES  CLASSIFIED  AS  TO  ANGLES.  A  triangle 
one  of  the  angles  of  which  is  a  right  angle  is  a  right 
triangle.  One  with  an  obtuse  angle  is  an  obtuse  trian- 
gle. If  all  the  angles  are  acute  the  triangle  is  an  acute 
triangle.  If  all  the  angles  are  equal  the  triangle  *is 
equiangular. 

A  ABC  is  right,  A  DEF  is  obtuse,  A  GHI  is  acute. 

Verify  with  a  protractor. 


20  PLANE  GEOMETRY 

62.  BASE  OF  A  TRIANGLE.    Unless  otherwise  stated, 
the  base  of  an  isosceles  triangle  is  the  side  which  is  not 
one  .of  the  two  equal  sides.    In  other  cases  the  base  of  a 
triangle  is  that  side  upon  which  it  is  conceived  to  stand. 

63.  VERTEX  ANGLE  OF  A  TRIANGLE.     The  angle  op- 
posite the  base  of  a  triangle  is  the  vertex  angle  and  the 
vertex  of  this  angle  is  the  vertex  of  the  triangle. 

64.  CONGRUENT  FIGURES.     Two  figures  that  can  be 
made  to  coincide  throughout  are  congruent.     Congru- 
ent means  of  the  same  shape  and  size,  while  equal  means 
of  the  same  size  only. 

1.  Draw  a  triangle  with  two  sides  2  and    2|    inches  respect- 
ively  and  included    angle   75°.      [30,    (3)].     Draw   another  with 
the  same  conditions.     Use  tracing  paper  to  compare  them  or  cut 
one  out  and  compare  them  by  superposition.     Are  they  congruent? 

2.  Draw  a  triangle  at  random.     Measure   two   sides  and  the 
included   angle.      Construct   another   using   the    measurements    ob- 
tained from  the  first  and  compare  them  by  superposition. 

3.  Draw  a  triangle  with  two  angles  of  85°  and  65°    respec- 
tively with   the  included   side  2   inches.     Draw  another  with   the 
same  conditions.     Compare  them  by  superposition.     What  is  the 
conclusion? 

4.  Draw  a  triangle  at  random.     Measure  two  angles  and  the 
included  side.      Construct    another   triangle    from   these    measure- 
ments and  compare  them  by  superposition. 

5.  Draw  a  straight  line    "1|-     inches  long.     With  the   dividers 
take  lines  respectively    f    and    1£    inches  and  from  the  ends  of 
the  first  line  describe  arcs  that  intersect.     Connect  the  point  of 
intersection  with  the  ends  of  the  first  line.     Take  the  same  three 
lines  in  any  order  and  construct  a  triangle.     Compare  these  tri- 
angles as  regards  congruence. 

6.  Construct   a   triangle   in  which   the   sides   are   2,   3,   and  4 
inches  respectively.       Construct  a  triangle  with  lines  of    l£,      If 
and  4  inches  respectively.      Try  again  with  lines  of  2,  2,  and  4 
inches  respectively ;  2,  3,  and  7 ;  4,  5,  and  6. 

7.  State  the  conclusions  of  Ex.  6  as  to  the  possibility  of  con- 
structing a  triangle  of  given  sides. 


RECTILINEAR  FIGURES  21 

PROPOSITION  I. 

65.  THEOREM.  Two  triangles  are  congruent  if 
two  sides  and  the  included  angle  of  the  one  are 
equal  respectively  to  two  sides  and  the  included 
angle  of  the  other.  A  A 


Given  A  ABC  and  &  A'B'C'  with  BA  —  B'Af, 
BC  =  B'C',  and  Z  B=  L  B' . 

To  Prove  A  ABC  =  A  A'B'C'. 

Proof .  Place  A  ABC  upon  A  A'B'C1  so  that  point  B 
will  fall  on  B'  and  BC  will  lie  along  B'C'. 

Then  C  will  fall  on  C'  for  BC  =  B'C'. 
BA  will  fall  along  B' A'  for  L  B  =  L  B'. 
A  will  fall  on  A"  for  BA  =  B'A', 

Since  C  falls  on  C'  and  J.  falls  on  A',  line  AC  will 
coincide  with  A'C'  for  only  one  straight  line  can  be 
drawn  through  two  given  points. 

•'•  A  ABC  =  A  A'B'C'  for  they  can  be  made  to  coincide. 

Therefore,  Two  triangles  are  congruent  if  two  sides 
and  the  included  angle  of  the  one  are  equal  respectively 
to  two  sides  and  the  included  angle  of  the  other. 

1.  To   find   the   distance   across   an   impassable  marsh:    Stake 
off  two   lines  AC  and  BD  that   cross   at  an  accessible  point   0. 
Measure  distances  OD  equal  to  OA  and  OB 

equal  to   OB.     Then  the   required    distance 

AB  is  equal  to  CD.     Why?  _ 

2.  In   the   school    grounds   or    elsewhere 
drive  three  stakes  for     A,  B,  and  0  of  ex- 


22  PLANE  GEOMETRY 

ample  1.      Make  the  remaining    measurements    and    test    the    ac. 
curacy    of   the    work    by  comparing  AB  and  CD. 

PROPOSITION  II. 

66.  THEOREM.  Two  triangles  are  congruent  if 
two  angles  and  the  included  side  of  the  one  are 
equal  to  two  angles  and  the  included  side  respec- 
tively of  the  other. 


Given  A  ABC  and  A  A' B'C'  with  /.B=/.B', 
BC  =  B'C',  LC=  L  C'. 

To  Prove  A  ABC  =  A  A'B'C'. 

Proof.  Place  A  ABC  upon  A  A'B'C'  so  that  B  shall 
fall  on  Br  and  BC  shall  fall  along  B'C'. 

Then  C  will  fall  on  C'  for  BC  is  given  equal  to  B'C'. 

BA  will  fall  along  B'A'  for  L  B=  L  B' . 

•'-A  will  fall  on  B'A'  or  B'A'  produced. 

Also  CA  will  fall  along  C'A'  for  L  C=  L  C'. 

•'-A  will  fall  on  C'A'  or  C'A'  produced. 

•'•  A  must  fall  on  A',  the  intersection  of  B'A'  and 
C'A'.  /WhyV) 

•'•  A^tgtT^  A  A'B'C'  for  they  can  be  made  to  coin- 
cide. 

Therefore,  Two  triangles  are  congruent  if  two  angles 
and  the  included  side  of  the  one  are  equal  to  two  angles 
and  the  included  side  respectively  of  the  other. 

67.  APPLICATION  OP  PROPOSITIONS  I  AND  II.  If  two 
triangles  are  congruent  the  angles  are  equal  respectively 


RECTILINEAR  FIGURES  23 

and  the  sides  are  equal  respectively.  It  is  important  to 
keep  this  in  mind  for  usually  the  purpose  in  proving 
triangles  congruent  is  to  prove  that  a  pair  of  angles  or 
line-segments  are  equal.  In  congruent  triangles  pairs  of 
equal  sides  lie  opposite  pairs  of  equal  angles  and  pairs 
of  equal  angles  lie  opposite  pairs  of  equal  sides.  The 
pairs  of  equal  parts  are  called  corresponding  or  homolo- 
gous parts. 

1.  Draw  a  triangle  at  random  and  with  a  straight  edge  and 
protractor; construct  a  second  triangle  having  two  sides  and  their 
included  angle  of  the  same  dimensions  as  the  corresponding  parts 
of  the  first.  Why  must  the  triangles  be  congruent?  Compare  the 
two,  using  tracing  paper,  as  a  test  for  accuracy  of  construction. 

PROPOSITION  III. 

68.     THEOREM.     //  two  sides  of  a  triangle  are 
equal,  the  angles  opposite  those  sides  are  equal. 
Given  A  ABC  with  AB  =  AC. 
To  Prove  Z  B  =  Z  C. 
Proof.      Draw   AM  to  represent  the 
bisector  of  Z  A  and  extend  it  to  meet 
B(1  as  at  M,     A.  B  and  C  will  be.  proved 
equal  if  it  can  be  shown  that  A  AMB 
and  AMC  are  congruent.    A  comparison 
of  the  two  triangles  shows 
AB  =  AC  by  hyp. 

Z  MAB  =  L  MAC  since  AM  bisects  Z  A. 
AM  is  common  to  the  two  triangles. 
•  '•A  AMB  =  A  AMC,    two    sides    and    the    in- 
cluded angle  of  the  one  being  equal  to  two 
sides  and  the  included  angle  of  the  other. 
Z  B  is  in  A  AMB  and  is  opposite  to  side  AM. 
Z  C  is  in  A  AMC  and  is  opposite  side  AM. 


24  PLANE  GEOMETRY 

•  '•  Z  B  =  L  C,  being  homologous  angles  in  con- 
gruent triangles. 

Therefore,  If  two  sides  of  a  triangle  are  equal,  the 
angles  opposite  those  sides  are  equal. 

REMARK.  Observe  that  LB  is  an  angle  in  each  of  two  tri- 
angles. In  A  AMB  it  lies  opposite  the  side  AM,  in  A  ABC  it 
lies  opposite  side  AC.  Likewise  LC  is  an  angle  in  each  of  two 
triangles. 

69.  COROLLARY.     An  equilateral  triangle  is  equiangu- 
lar. 

Proof.    Use  proposition  III. 

PROPOSITION  IV. 

70.  THEOREM.     The  bisector  of  the  vertex  an- 
gle of  an  isosceles  triangle  bisects  the  base  and  is 
perpendicular  to  the  base. 


**  C 

Given  A  ABC,  with  AB  =  AC  and  Z  MAB  =  Z  MAC. 
To  Prove  BM  =  MC  and  AM  LBC. 
Proof.     A  comparison  of  the   A  AMB  and   A  AMC 
shows 

AB  =  AC  by  hyp. 
Z  MAB  =  £MAC  by  hyp. 
Z  B  —  Z  (7,  being  opposite  equal  sides  of  the 

isosceles  A  ABC. 

•  '•A  AMB  E  A  AMC,  having  two  angles  and 
the  included  side  of  the  one  equal  to  the  cor- 
responding parts  of  the  other. 


RECTILINEAR  FIGURES  25 

In  A  AMB  side  BM  is  opposite  Z  MAB  and 

in  A  MAC  side  M.C  is  opposite  L  MAC. 
-'•BM  =  MC,  being  homologous  sides  in  con- 
gruent triangles. 

Also   Z  AMB  =  Z  AMC,  being  homologous  an- 
gles in  congruent  triangles. 
•'•A.  AMB  and  AMC  are  right  angles.     Why? 
•'•AM±BC.     Why? 

Therefore,  The  bisector  of  the  vertex  angle  of  an 
isosceles  triangle  bisects  the  base  and  is  perpendicular 
to  the  base. 

71.  DISTANCE.     The  length  of  the  sect  joining  two 
points  is  the  distance  between  these  points. 

It  is  to  be  noted  that  distance  is  always  measured  on 
the  shortest  line. 

72.  PERPENDICULAR     BISECTOR.       A     perpendicular 
erected  at  the  mid-point  of  a  line-segment  is  the  per- 
pendicular bisector  of  that  segment, 

PROPOSITION  V. 

Z3.  THEOREM.  Any  point  in  the  perpendicular 
bisector  of  a  line-segment  is  equidistant  from  the 
extremities  of  the  segment. 


B    ^  M  O 

Given  segment  BC  with  BM  —  MC,  MN  1  BC  and  A 
any  point  in  MN. 
To  Prove  AB  =  AC. 


26 


PLANE  GEOMETRY 


Proof.     SUG.     If  A  BMA  and  A  CM  A  can  be  proved 
congruent,  segments  AB  and  AC  will  be  seen  to 
be  equal.     Search  the  conditions  given  to  see  if 
there  are  enough  elements  equal  to  make  the  tri- 
angles congruent  by  Prop.  I  or  II.    See  method  of 
§  70.    Compare  AB  and  AC,  giving  authorities. 
74.     PARTS  OF  A  THEOREM.     From  the  propositions 
studied  thus  far  it  is  seen  that  in  a  theorem  certain  con- 
ditions are  given  and  a  certain  thing  is  to  be  proved. 
The  conditions  given  constitute  the  hypothesis  and  the 
truth  to  be  established  is  the  conclusion.     The  course  of 
reasoning  by  means  of  which  the  conclusion  is  estab- 
lished is  the  proof  or  demonstration.     Each  statement 
in  a  proof  must  ,be  justified  by  reference  to  the  hypoth- 
esis or  to  definition,  axiom,  postulate,  or  former  prop- 
osition. 

1.  Two  right  triangles  are  congruent  if 
the  sides  of  the  right  angle  of  the  one  are 
equal  respectively  to  the  sides  of  the  right 
angle  of  the  other. 

2.  Lines  AB  and  CD  bisect  each  other 
at  0. 

Prove  A  AOC  =  A  BOD  and  AC  =  BD.  o 

3.  OC   bisects   L  AOB   and   DELOC 
at  F. 

Prove     A  OFD  -    A   OFE,   OD  =  OE, 
and  L  ODF  -  Z  OEF. 

4.  In   A  ABC,  AB  =  AC  and  Z  BAM  = 
L  CAN.        Prove      A  BAM  =  A  CAN      and 
&AMN  is  isosceles. 

5.  In    A  ABC,    AB  =  AC    and    D,   E,   F 
are  the  mid-points  of  AB,  BC,  and  CA  re- 
spectively.    Prove  A  DEF  isosceles. 


RECTILINEAR  FIGURES 


27 


PROPOSITION  VI. 

75.  THEOREM.  Two  triangles  are  congruent  if 
the  sides  of  the  one  are  equal  respectively  to  the 
sides  of  the  other. 


Given  A  ABC  and  A  A'B'C',  AB  =  A'B',  BC  =  B'C 
and  CA  =  C' A' . 
To  Prove  A  ABC  E  A  A'B'C'. 

SUG.  1.  The  conclusion  follows  at  once  if  it 
can  be  shown  that  Z  A  =  L  A',  Z5=Z5', 
or  Z  C  =  L  C'. 

2.  Place  A  A'B'C'  so  that  a  side,  as  A'C' 
coincides  with  its  equal  AC,  the  two  points  B  and 
B'  being  on  opposite  sides  of  AC. 

3.  Draw  BB'. 

4.  Z  ABB'  =  L  AB'B.    Why  ? 

5.  Z  055 '  =  Z  05 '  5.    Why  ? 

6.  •'•  Z  ABC  =  Z  A5'C.    Why  ? 

7.  •'•  A  A50  =  A  A'B'C'.     Why? 
Demonstrate  the  theorem  again,  placing  a  second  pair 

of  equal  sides  in  coincidence. 
Therefore— 

1.  The    straight   lines   which    bisect    the 
equal    angles    of   an    isosceles    triangle    and 
terminate  in  the  opposite  sides  are  equal. 

2.  A  ABC   is    equilateral    and   AD  =  BE 
=  CF.     Prove  A  DEF  equilateral. 


28  PLANE  GEOMETRY 

PROPOSITION  VII. 

76.  THEOKEM.     Any  point  ivhich  is  equidistant 
from  the  extremities  of  a  line-segment  is  in  the 
perpendicular  bisector  of  the  segment. 

The  figure  is  to  be  constructed  by  the  pupil. 
Given  the  segment  AB  with  mid-point  C,  and  D  any 
point  such  that  DA  —  DB  and  ME  the  line  through  C 
and  D. 
To  Prove  ME  1  AB  at  C. 

SUG.     Compare  &  ACD  and  BCD  and  also  the 
two  adjacent  angles  at  C. 

77.  COROLLARY.     Any    two    points   each    equidistant 
from  the  extremities  of  a  line-segment  determine  the  per- 
pendicular bisector  of  the  segment. 

SUG.     How    many    points    determine    a    line? 
§50  (1). 

1.  Nail  together  with  one  nail  at  each  vertex  three  narrow 
strips  of  wood  so  as  to  form  a  triangle.  Can  it  be  racked  out  of 
shape  without  loosening  the  joints?  Why? 

.  2.     Remove  the  bottom  from  a  chalk  box.     Can  the  box  now  be 
racked   out    of   shape?      Nail   a    strip    across   two    adjacent   sides. 
Can  it  still  be  racked  out  of  shape?     Give 
authorities   for  your   answers. 

3.  Why  does  the  stay-lath  AB  which  the 
carpenter    nails    across    a    window    or    door 
frame  hold  it  in  a  fixed  shape? 

4.  Observe  at  the  next  opportunity  the 
board  nailed   obliquely  across   the   studding 
as  they  are  erected  in  putting  up  the  frame 
of   a  building  or   the    diagonal    strip    on  a 
gate,   as  AB  in  the  figure. 

5.  Name  at  least  five  other  applications  of  prop. 
VI  for  insuring  stability  in  mechanical  constructions. 

6.  If    the    perpendicular   bisector    of    the    base    of    a   triangle 
passes  through  the  vertex,  the  triangle  is  isosceles.  Use  fig.  of  8  70. 


RECTILINEAR  FIGURES  29 

PROPOSITION  VIII. 

78.     PROBLEM.     Construct  a  triangle  when  the 
three  sides  are  given.  ^ 


Given  sides  of  a  triangle  as  a,  6,  c. 
To  Construct  the  triangle. 

SUG.     1.     Draw  a  line  and  with  dividers  lay  off 
upon  it  one  of  the  sides  as  AB  =  c. 

2.  With  the  dividers  find  a  point  C  that 
is  the  distance  a  from  B  and  the  distance  6  from 
A.  §  50  (8).  A  ABC  is  the  required  triangle. 

PROPOSITION  IX. 

79.     PROBLEM.     Construct  upon  a  given  ray  an 
angle  equal  to  a  given  angle. 


E 

Given  ray  AB  and  ZC. 

To  Construct  an  Z  upon  AB  equal  to  Z  C  with  its 
vertex  at  A. 

SUG.     1.     With    dividers   measure    equal    dis- 
tances as  CD  and  CE  on  the  sides  of  Z  C. 

2.  Find  F   on  AB  so  that  AF  =  CE. 

3.  With  the  dividers  find  point  X  so 
that  AX  =  CD  and  FX  =  ED. 

4.  Then   Z  XAF=  Z  DCE  or   Z  C. 
For  the  4  AXF  and  CDE  have  the  three  sides  re- 
spectively equal. 


30  PLANE  GEOMETRY 

PROPOSITION  X. 

80.     PROBLEM.     Construct  a  triangle  congruent 
to  a  given  triangle. 


Given  A  ABC. 

A 

To  Construct  a  A  congruent  to  A  ABC.    Make  three 
constructions. 

SUG.     1.     Use  the  problem  §  78. 

2.  Use  §  65. 

3.  Use  §  66. 

In  these  problems  use  a  straight  edge  and  dividers  only. 

1.  Prove   the   theorem    of    §  75,   using   two   scalene    triangles 
with  a  pair  of  the  shorter  sides  coincident. 

2.  Upon  a  given  segment  as  a  base  construct  an  equilateral 
triangle. 

3.  Construct  the  perpendicular  bisector  of   a  given  line-seg- 
ment.    Sug.  §  77.  P 

4.  Let    0    be    a    given    point    in    AB. 
Construct  the  perpendicular  to  AB  through  0. 


Sue.     If  P  be  a  point  in  this  perpendicular,  how  may  P 
be  obtained?     Why  must  line  OP  when  thus  determined  be 
the  required  perpendicular. 
o<     Solve    example    4    assuming    that 
O  is  outside  AB. 

6.  Required    to    bisect    a    given    an- 
gle, Z  ABC. 

Sue.     Obtain   a   point  P  such  that   when  BP  is  drawn, 
Z  DBF  =  Z  EBP.     §  75.  A 

7.  To    measure    a    distance    AB    across 
a  river.     From  A  take  any  convenient  line, 
AC,  of  known  length,  measure  angles  A  and 

C.     In    some   convenient   place    construct    a         C1 

A    congruent    to    A  ABC    and    measure    the  side  corresponding  to 

AB.     State  the  authority  for  the  conclusion.    Sug.  $  60. 


RECTILINEAR  FIGURES  31 

PROPOSITION  XL 

81.     THEOREM.     Only  one  perpendicular  can  be 
drawn  from  a  given  point  to  a  given  line. 


M\ 
\ 


\ 
\ 
\ 


Given  line  CD  with  A  a  point  not  in  CD,  AO  1  CD, 
and  AM  any  other  line  from  A  to  CD. 
To  Prove  AM  not  perpendicular  to  CD. 

SUG.     1.     Produce  AO  to  B,  making  OB  =  OA. 
Connect  M  and  B. 

2.  Since  AOB  is  a  straight  line  by  con- 
struction, AMB  is  not  a  straight  line.    Why? 

3.  •*•  Z  AMB  is  not  a  straight  angle. 

4.  Compare  A  AOM  and  A  BOM.  Auth. 

5.  Compare  Z  ^.ifO  with  Z  53fO. 

6.  •'•  Z  ^MO  is  one  half  of  Z  AM£. 

7.  •'•  Z  J.MO  cannot  be  a  right  angle. 

8.  •'•  AM  is  not  perpendicular  to  CD. 

9.  •'•  AO  is  the  only  perpendicular  from 
A  to  CD,  since  J.M  represented  all  other  lines 
from  A  to  CD. 

82.  HYPOTENUSE.  The  side  of  a  right  triangle 
opposite  the  right  angle  is  the  hypotenuse.  The  other 
sides  are  usually  called  the  sides  or  legs. 

1 .  Upon  a  given  segment  as  a  base  construct  an  isosceles  tri- 
angle. 


32 


PLANE  GEOMETRY 


PROPOSITION  XII. 

83.  THEOREM.  Two  right  triangles  are  con- 
gruent if  the  hypothenuse  and  an  adjacent  angle 
of  the  one  are  equal  respectively  to  the  hypothe- 
nuse  and  an  adjacent  angle  of  the  other. 


Given  A  ABC  and  A  A'B'C'  with  L  B  and  LE'  right 
angles,  AC  =  A'C',  and  Z  C=Z  C". 

To  Prove  A  ABC  =  A  A'B'C'. 

Proof.  Place  A  ABC  upon  A  A'B'C'  so  that  A  falls  on 
A  and  AC  falls  along  A'C".     Where  will  C  fall  and  why  ? 

Also  C£  will  fall  along  C'B'  for  Z  C=  L  C'. 

B  will  fall  on  C'Bf  or  C"£'  produced.     Why? 

Since  A  falls  on  A'  and  £<7  falls  along  B'C',  AB  is  the 
perpendicular  from  A'  to  B'C". 

But  A'B'  is  by  hyp.  perpendicular  to  B'C'.    Why? 
AB  must  fall  along  A'B'.     Why  ? 

•'•5  falls  on  B'.     Why? 
•  A  ABC  =  A  A'B'C'.     Why? 

Therefore — 

1.  If  two  isosceles  triangles  have  their  bases  coincident  and 
their  vertices  on  opposite  sides  of  the  common  base,  prove: 

(1)  The   entire   figure    is   divided  into    two   congruent   tri- 
angles by  a  line  connecting  the  opposite  vertices. 

(2)  This  line  is  the  perpendicular  bisector  of  the  common 
base. 

2.  If  in  example  1  the  vertices  are  assumed  to  be  on  the  same 
side  of  the  common  base,  are  the  above  conclusions  true? 


RECTILINEAR  FIGURES  33 

PROPOSITION  XIII. 

84.  THEOREM.  Two  right  triangles  are  con- 
gruent if  the  hypothenuse  and  a  side  of  the  one 
are  equal  respectively  to  the  hypothenuse  and  a 
side  of  the  other. 


^C      B 


Z  B' 


pro- 


Given  A  ABC  and   AA'B'C',  with   Z5  and 
right  angles,  AC  =  A'C',  and  AB  =  A'B'. 
To  Prove  A  ABC  =  A  A'B'C'. 
Proof.    SUG.  1.     The    triangles    are    congruent 
vided  Z  C=  /.  C'  or  /.  A=  £  A' .    §  83. 

2.  Place  the  triangles  so  that  A'  falls 
on  J.,  A'R'  along  J.#,  and  C  and  C'  on  opposite 
sides  of  AB. 

3.  Where  will  B'  fall  and  why? 
OBC'  is  a  straight  line.     Why? 
Figure  ACC'  is  a  triangle.    Why? 
What  kind  of  a  triangle  is  ACC'  ? 

Why? 

7.  Compare   Z  C  and  Z  C".     Auth. 

8.  Compare  A  ABC  and  J/5'C".   Auth. 

Therefore— 

Yvrhy  is  it  necessary  to  prove  CBC'  a  straight  line? 

85.  COR.  The  perpendicular  from  the  vertex  to  the 
base  of  an  isosceles  triangle  bisects  the  ~base  and  the  ver- 
tex angle. 


4. 
5. 

6. 


34  PLANE  GEOMETRY 

86.  PARALLEL  LINES.     Straight  lines  that  lie  in  the 
same  plane  and  cannot  meet  however  far  produced  are 
parallel  lines. 

87.  POSTULATE.     Through  a  given  point  only  one  line 
can  be  drawn  parallel  to  a  given  straight  line. 

PROPOSITION  XIV. 

88.  THEOREM.     Two  lines  in  the  same  plane 
and  perpendicular  to  the  same  line  are  parallel. 

M 


Given  AB  and  CD  in  the  same  plane  and  each  per- 
pendicular to  MN. 

To  Prove  AB  II  CD. 

Proof.  Suppose  AB  and  CD  ngt  parallel,  i.e.  sup- 
pose they  meet  if  sufficiently  produced.  Then  from  this 
point  of  meeting  there  will  be  two  perpendiculars  to  the 
line  MN,  viz.  AB  and  CD.  But  this  is  impossible  for 
only  one  perpendicular  can  be  drawn  from  a  given  point 
to  a  given  line.  §  81. 

Hence  the  supposition  that  AB  and  CD  meet  is  false. 

•'•  AB  ||  CD,  for  they  are  in  the  same  plane  and  do  not 
meet. 

Therefore— 

89.  INDIRECT  PROOF.  The  proof  in  §  88  is  an  example 
of  indirect  proof.  In  this  method  the  theorem  is  assumed 
to  be  not  true  and  the  assumption  is  then  shown  to  be 
false  in  that  it  leads  to  the  contradiction  of  known  facts. 
§  90  is  another  example  of  indirect  proof. 


RECTILINEAR  FIGURES  35 

1.  The  perpendiculars  from  the  extremities  of  the  base  of  an 
isosceles  triangle  to  the  opposite  sides  are  equal. 

2.  Discuss  exercise  1,  when  the  vertex  angle  is  obtuse;  when 
it  is  right. 

3.  The  perpendiculars  drawn  from  the  mid-point  of  the  base 
of  an  isosceles  triangle  to  the  other  sides  are  equal. 

4.  In   &ABC,  D  is  the   mid-point   of   BC.     BE  and   CF  are 
perpendiculars  drawn  from  B  and  C  respectively  to  AD  (AD  be- 
ing produced  through  D).     Prove  BE  =  CF. 

5.  Discuss  exercise  4,  if  A  ABC  is  isosceles  with  AB  =  AC. 

PROPOSITION  XV. 

90.  THEOREM.  //  one  of  two  parallel  lines  is 
perpendicular  to  a  given  linef  the  other  is  per- 
pendicular to  the  same  line. 


~~G 

N 

Given  AB  \\  CD,  AB  1  MN  at  E  and  CD  intersecting 
MN  at  F. 

To  Prove  CD  1  MN. 

Proof.  Suppose  that  CD  is  not  perpendicular  to  MN. 
Let  HGr  represent  the  line  which  is  perpendicular  to  MN, 
at  F. 

Then  HGllAB.    Why? 

But  CD  II  AB.    Hyp. 

Then  through  F  there  are  two  lines  parallel  to  AB. 

This  contradicts  the  postulate,  §  87. 

•'•  the  supposition  that  CD  is  not  perpendicular  to  MN 
is  false.  -'-CD1.MN. 

Therefore — 


36  PLANE  GEOMETRY 

91.  DISCUSSION.     Observe  that  HG  is  given  the  prop- 
erty which  CD  is  assumed  not  to  possess.    This  is  legiti- 
mate since  it  is  known  that  through  point  F  there  is  a 
line  perpendicular  to  MN  and  if  CD  is  not  this  perpen- 
dicular we  may  represent  it  by  HG. 

DIRECT  PROOF.  Without  making  any  supposition  as  to 
CD,  draw  a  perpendicular  to  MN  through  F.  Then  as 
before  both  HG  and  CD  are  parallel  to  AB.  Whence 
HG,  if  correctly  drawn,  must  coincide  with  CD  for  other- 
wise there  would  be  two  lines  through  a  given  point 
parallel  to  the  same  line. 

Since  HG  1  MN  by  construction  and  CD  coincides 
with  H G,  then  CD  1  MN. 

92.  TRANSVERSAL.     A  line  that  cuts  two  or  more  lines 
is  a  transversal  or  secant  line. 

93.  CLASSIFICATION  OF  ANGLES   MADE  BY  A   TRANS- 
VERSAL.    The  angles  made  by  two  lines  and  a  trans- 
versal are  classified  as  follows: 

1.  The   angles  1,   2,   3  and  4  are 
interior  angles. 

2.  The  angles  5,  6,  7  and  8  are 
exterior  angles.  c- 

3.  The  pairs  1,  4  and  2,  3  are  al- 
ternate-interior angles.  F 

4.  The   pairs  5,   8   and  6,   7  are  alternate-exterior 
angles. 

5.  The  pairs  (1,  7),  (2,  8),  (6,  4),  and  (5,  3)  are 
corresponding  anglesW 

1.  Any  point  in  the  bisector  of  an  angle  is  equidistant  from 
the  sides  of  the  angle.  ^^ 

Suo.    Draw  perpendiculars  from  a  point  l^me  bisector  to 
the  si^s  apd  prove  them  equal. 


RECTILINEAR  FIGURES  37 

PROPOSITION  XVI. 

94.  THEOREM.  //  two  parallel  lines  are  cut  by 
a  transversal  the  alternate-interior  angles  are 
equal.  M  G^*  p 


F' 

Given  AB  I!  CD  and  cut  by  the  transversal   EF  at 
points  G  and  H  respectively. 
To  Prove  Z  AGF=  Z  DHE  and   Z  BGF=£  CHE. 

SUG.     1.     Through   0,  the   mid-point   of   GH, 
draw  ON  1  CD  and  extend  it  to  AB  at  M. 

2.  What  relation  does  J/N  sustain  to 
AB  and  why? 

3.  Compare      A  OMG     and      A  ONH. 
Auth. 

4.  Compare      /.  AGF      and      Z  Z>##. 

5.  Compare  Z  £<7,F  and  Z  CHE. 
Therefore— 

REMARK.  In  answering  Sug.  4  observe  that  it  is  not  known 
that  OM  and  ON  are  equal.  How  then  can  it  be  shown  that 
Z  AGF  and  Z.  DHE  are  homologous  angles  in  congruent  triangles? 

1.  Any  point    equidistant    from   the   sides    of    an   angle    is   in 
the  bisector  of  the  angle. 

2.  Any  point   not  in  the   bisector   of 
an    angle    is    not    equidistant    from    the 
sides. 

Sue.      Draw  PM  1  BC  and  PN  1  BA. 
Let  PN   meet   the   bisector   in   0.    Drato 
OG1.BC.     Then   OG  =  ON.     Why? 
PG  <  PO  +  OG.     •'•  PG  <  PN.     PM  <  PG.     •'•  PM  <  PN. 


Give  the  i^^n  for  each  of  these  steps. 

3.     Demonstrate  Ex.   2    by  the  indirect  method.  ^ 


38  PLANE  GEOMETRY 

The  pupil  is  expected  to  make  detailed  proofs  of  the 
following  three  corollaries. 

95.  COR.     //  two  parallel  lines  are  cut  by  a  trans- 
versal, the  corresponding  angles  are  equal. 

96.  COR.     //  two  parallel  lines  are  cut  by  a  trans- 
versal, the  alternate-exterior  angles  are  equal. 

97.  COR.     //  two  parallel  lines  are  cut  by  a  trans- 
versal, the  interior  angles  on  the  same  side  of  the  trans- 
versal are  supplementary. 

PROPOSITION  XVII. 

98.  THEOKEM.     //  two  lines  in  the  same  plane 
are  cut  by  a  transversal  and  the  alternate-interior 
angles  are  equal,  the  lines  are  parallel. 


Given  AB  and  CD  cut  by  the  transversal  EF  at  G  and 
H  respectively  with  Z  AGF  =  L  DEE. 
To  Prove  AB  II  CD. 

SUG.  1.  Suppose  AB  not  parallel  to  CD  and 
let  MN  represent  that  line  through  G  which  is 
parallel  to  CD. 

2.  Compare  Z  MGF  with  Z  DHE.  Auth. 

3.  Compare  Z  AGF  with  Z  DUE.  Auth. 

4.  Compare  Z  MGF  with  Z  AGF.  Auth. 

5.  Compare  this  conclusion  with  the  facts 

6.  What  of  the    supposition    that  AB  is 
not  parallel  to  CD? 

1.    What  is  the  true     relation    between 
AB  and  CD? 


EECTILINEAR  FIGURES  39 

Therefore— 

Prove  this  proposition  by  the  direct  method.    See  §  91. 

REMARK.  The  second  of  the  above  figures  will  be  useful  if 
chalks  of  different  colors  are  used  to  represent  the  lines  AB  and 
MN. 

99.  COR.  I.    //  two  lines  in  the  same  plane  are  cut  by 
a  transversal  and  the  corresponding  angles  are  equal, 
the  lines  are  parallel. 

100.  COR.  II.    //  two  line®  in  the  same  plane  are  cut 
l)y  a  transversal  and  the  interior  angles  on  the  same  side 
of    the    transversal   are    supplementary,    the    lines   are 
parallel. 

101.  CONVERSE.     If  two  propositions  are  so  related 
that  a  condition  of  the  first  is  the  conclusion  of  the  sec- 
ond and  the  conclusion  of  the  first  is  a  condition  of  the 
second,  each  proposition  is  called  the  converse  of  the 
other.    For  example,  Prop.  XVII  is  the  converse  of  Prop. 
XVI  for  in  the  first  of  these  the  hypothesis  is  ^1 B  II  CD 
and  the  conclusion  is  Z  AGF=  Z  DEE;  in  the  second 
the  hypothesis  is  Z  AGF—  L  DEE  and  the  conclusion 
is  AB  II  CD.  It  is  to  be  noted  that  there  is  one  condition 
common  to  the  two  propositions,  viz:  two  lines  and  a 
transversal  are  given : 

The  converse  of  a  true  statement  is  not  always  true. 
For  example,  all  right  angles  are  equal  but  not  all  equal 
angles  are  right  angles. 

1.  If  AB\\CD  and  x  =  70°,  how  many      A /       g 

degrees  in  each  one  of  the  various  angles  in  y/£ 

the  figure?  ./  T> 

2.  If  AB  ||  CD  and  y — x  =  80°,  how  many  degrees  in  each  of 

the    various    angles?       How    many    if    »s  rl(f  JL.f  .'^£f  :Slff 

7835 


40  PLANE  GEOMETRY 

3.  Two  angles  are  complementary  and  their  difference  is  38°. 
How  many  degrees  in  each? 

4.  Given  a  line  AB  and  a  point  C  not  on  AB.     Construct  a 
line  through  C  and   parallel  to  AB.     Sug.     Try  to  reproduce  the 
conditions  of  Prop.  XVII  or  of  one  of  its  corollaries. 

PROPOSITION  XVIII. 

102.  THEOREM.  Two  angles  the  sides  of  which 
are  respectively  parallel  and  extend  in  the  same 
or  in  opposite  directions  from  the  vertices  are 
equal. 


Given  A  A,  B,  B'  with  AC  I!  BE,  AD  II  BF  and  ex- 
tending in  the  same  directions  from  the  vertices  A  and 
B;  and  with  AC  il  B'E',  AD  II  B'F'  and  extending  in 
opposite  directions  from  the  vertices  A  and  B'. 
To  Prove  L  A  =  L  B  =  L  B'. 

SUG.     1.     Extend    AD    and    B'E'    until   they 
meet  at  0.     (Why  must  they  meet?) 

2.  Compare   A  A  and  B' '. 

3.  Extend  AD  and  BE  until  they  meet 
at  M. 

4.  Compare  /i  A  and  B. 
Therefore—- 
in what  kind  of  triangles  does  the  bisector  of  the  vertex  angle 

coincide  with  the  perpendicular  from  the  vertex  to  the  base  and 
with  the  median  to  the  base  I  The  line  from  the  vertex  to  the 
mid-point  of  the  base  is  called  the  median  to  the  base. 


RECTILINEAR  FIGURES  41 

103.  EXTERIOR   ANGLE.    An  angle  formed  A  B 
by  one  side  of  a  rectilinear  figure    and    an     ^Ssv£ 
adjacent   side   extended   is   an   exterior  angle.     In  the 
figure  n  is  an  exterior  angle  of  the  triangle  ABC. 

104.  OPPOSITE  INTERIOR  ANGLE.    In  the  above  figure 
with  respect  to  the  exterior  angle  n,  A  and  C  are  the 
opposite  interior  angles. 

PROPOSITION  XIX. 

105.  THEOREM.     An  exterior  angle  of  a  trian- 
gle is  equal  to  the  sum  of  the  opposite  interior 
angles. 


»*-  — c 

Given.     A  ABC  with  exterior  angle  DAC. 
To  Prove  Z  DAC=  Z  B  +  L  C. 

SUG.     1.     Through    the    vertex   A    draw    line 
MN  \\  BC. 

2.  Compare  Z  p  with  Z  B.     Auth. 

3.  Compare  Z  q  with  Z  C.     Auth. 

4.  Complete  the  proof. 
Therefore— 

1.  Two  lines  in  the  same  plane,  each  parallel  to  a  third  line 
in  that  plane,  are  parallel.     Sug.     Draw  a  transversal,  a  line  per- 
pendicular to  the  third  line,  or  use  the  indirect  method. 

2.  Two   lines   perpendicular   to   two   intersecting  lines   respec- 
tively cannot  be  parallel.     Sug.     Use  indirect  proof. 

3.  Two  angles  are  supplementary  and  their  difference  is  65°. 
Find  the  angles. 

4.  State   all  methods   thus  far   given   for   proving   two   lines 
parallel. 


42  PLANE  GEOMETRY 

PROPOSITION  XX. 

106.     THEOKEM.     The  sum  of  the  interior  an- 
gles of  a  triangle  is  equal  to  two  right  angles. 


D * 


Given  A  ABC. 

To  Prove  ZA  +  Z£+ZC  =  2rt.  A. 
Sue.     1.     Extend  one  side  as  CB. 

2.     Compare  Z  m  +  Z  n  with 
Z  «,  +  Z  4  +  Z  C. 

Therefore— 

107.  COR.     A  triangle  can  have  at  most  one  obtuse 
or  one  right  angle. 

108.  COR.     The  acute  angles  of  a  right  triangle  are 
complementary. 

1.  Draw  through  B  of  the  figure   of   §  105   a   line   parallel  to 
AC  and  prove  the  theorem. 

2.  Extend   side    CA    through   A    to   a   point   E   and   prove   the 
theorem  using  the  angles  p,  EAD,  RAM. 

3.  Make   the  constructions  of  exercises   1   and   2   for  exterior 
angles  at  B  and  C  and  prove  the  theorem. 

4.  The  bisector  of  one  of  two  vertical  angles  is  the  bisector 
of  the  other. 

5.  The    bisectors    of    two    vertical    angles    are    in    the    same 
straight    line.      Sug.      Prove    that    the   bisectors    form    a    straight 
angle. 

0.     The    bisectors   of   two    supplementary    adjacent    angles    an- 
perpendicular  to  eacn  other. 


RECTILINEAR  FIGURES  43 

PKOPOSITION  XXI. 

109.  THEOREM.     //  two  triangles  have  two  an- 
gles of  the  one  equal  respectively  to  two  angles  of 
the  other,  the  third  angles  are  equal. 

Given  A  ABC  and  A  A'E'C'  with   L  A  =  L  A'  and 
Z£=Z£', 
To  Prove  ZC=ZC'. 
SUG.    Use  §  106. 

110.  COR.     //  two  right  triangles  have  an  acute  angle 
of  the  one  equal  to  an  acute  angle  of  the  other,  the  re- 
maining angles  are  equal. 

1.  How  many  degrees  in  each  angle  of  an  equilateral  triangle? 

2.  Construct  an  angle  of  60°;  an  angle  of  30°;  an  angle  of 
45°.  Use  dividers  and  straight  edge. 

3.  Construct  a  right  triangle  with  acute  angles  of  60°   and 
30°. 

4.  Trisect  a  right  angle. 

As  a  general  problem  the  trisection  of  an  angle  is  impossible 
by  the  use  of  dividers  and  straight  edge  only.  The  pupil  should 
note  that  this  example  is  a  particular  case  of  the  unsolved  general 
problem. 

5.  How  many  degrees  in  each  angle  at  the  base  of  an  isosceles 
right  triangle? 

6.  The  vertex  angle  of  an  isosceles  triangle  is  34°  40'.     How 
many  degrees  in  each  angle  at  the  base  of  the  triangle! 

7.  An  exterior  angle  at  the  base  of  an  isosceles  triangle  is 
100°.     Find  each  angle  of  the  triangle. 

8.  The  angles  of  a  triangle  are  5z°,   25x°,  and  SO0.     Find 
the  unknown  angles. 

9.  The  acute  angles  of  a  right  triangle  are  x°  and  8#°.    Find 
them. 

10.  The   difference  between   the  acute  angles  of  a  right  tri- 
angle is  26°.     Find  them. 

11.  An  exterior  angle  to  one  of  the  acute  angles  of  a  right 
triangle  is  140°.    Find  the  angles  of  the  triangle. 


44 


PLANE  GEOMETRY 


PROPOSITION  XXII. 

111.  THEOREM.  //  two  angles  of  a  triangle  are 
equal,  the  sides  opposite  are  equal  and  the  triangle 
is  isosceles. 


Given  A  ABC  with  L  B  =  L  C. 
To  Prove  AB  =  AC. 


SUG. 

as  AM. 

Auth. 

Auth. 


1.  Drop  a  perpendicular  from  A  to  BC, 

2.  Compare      Z  BAM     and      L  CAM. 

3.  Compare      A  BAM     and      A  CAM. 

4.  Compare  AB  and  AC. 


Therefore — 

1.  The    vertex    angle    of    an    isosceles    triangle    is    i    o:?    an 
angle  at  the  base.    Find  the  angles  of  the  triangle. 

2.  The  angles  of  a  triangle  are  denoted  by  3#,  7x,  and  5x. 
Find  each  angle. 

3.  Two  right  triangles  are  congruent  if  a  side  and  the  oppo- 
site angle  of   the  one  are   equal  respectively  to  a  side  and  the 
opposite  angle  of  the  other. 

4.  Perpendiculars  to  the  sides   drawn   from   any  point  in  the 
base  of  an  isosceles  triangle  make  equal  angles  with  the  base. 

5.  Construct   a  protractor   of  wood   or 
thick   cardboard  having  a  six   to   nine  inch 
radius.      To   do    this,    place   the    small    pro- 
tractor upon  the  material  to  be  used  and  ex- 
tend  the    radii   to    the   required   length.     Fasten    a   pendulum   or 
marker  at  the  center.     This  can  be  used  to  take  vertical  angles. 


RECTILINEAR  FIGURES 


(5.     The   acute  angles  of  a  right  triangle  have  the   ratio   2:5. 
Find  them. 

7.  The  angles  of  a  triangle  are  x,  y,  s,  with  x  +  y  =  80°  and 
z —  y  —  50°.     Find  x,  y,  z. 

8.  To    find    the    height    of    a    flag  B 
staff,    measure    back    from    the    base    of 

the  staff  on  level  ground  a  convenient 
distance  as  AC.  From  C  sight  along  the 
straight  edge  of  the  protractor  to  B. 
The  marker  swinging  freely  will  indi- 
cate upon  the  protractor  the  angle  at  B.  (Why?)  Then  find  ZC 
(How?)  and  on  some  convenient  place  construct  a  rt.  triangle  hav- 
ing the  parts  B,  C,  AC,  and  measure  the  side  homologous  to  AB. 

9.  To   find  the  height  of  a  tower  when  the  point   directly 
under  the  top  of  the  tower  cannot  be  obtained.    Measure  Z  m  at 
eome  convenient  point  on  level  ground  and  then  find  Z  n.     Meas- 
ure a  distance  CD,  the  line  BCD  being  straight,  and  at  J>  deter- 
mine  the   angle.     On   convenient   ground 

construct  a  triangle  with  the  given  parts 
Z  n,  L  D,  and  side  CD.  Extend  the  side 
DC  until  it  meets  the  perpendicular 
dropped  to  it  from  the  point  correspond- 
ing to  A.  The  length  of  this  perpendicu- 
lar is  the  height  of  the  tower. 

PROPOSITION  XXIII. 

112.     THEOREM.     Any  side  of  a  triangle  is  less 

than  the  sum  of  the  two  other  sides  and  greater 

than  their  difference. 

Given  A  ABC. 

To  Prove  AB  <  BC  +  CA  and 
AB>BC~  CA. 

SUG.     1.     AB<BC  +  CA.    Why? 
AB  +  CA>BC.     Why? 


1. 
2. 
3. 


•'.AB>BC  —  CA.  §  49  (4). 
Therefore— 

1.     Two   sides   of   a  triangle  are   6   ft.   and  8   ft.  respectively. 

What  are  the  limits  to  the  length  of  the  third  side? 


46  PLANE  GEOMETRY 

PROPOSITION  XXIV. 

113.  THEOREM.  //  two  angles  of  a  triangle  are 
unequal,  the  sides  opposite  these  angles  are  un- 
equal and  the  side  opposite  the  greater  angle  is 
the  greater. 


Given  A  ABC  with  L  C>  L  B. 
To  Prove  AB  >  AC. 

SUG.     1.     Draw    CM  from    C  to   AB  so   that 
L  BCM  =  L  B.    Show  that  this  is  possible. 

2.  M  must  fall  between  A  and  B.    Why? 

3.  Compare  CM  +  MA  with  AC.    Auth. 

4.  Compare  CM  and  BM.    Auth. 

5.  Complete  the  proof. 
Therefore — 

114.  COB.  1.       The  hypotenuse  is  the  longest  side  of 
a  right  triangle. 

115.  COR.  2.     The  perpendicular  from  a  point  to  a 
line  is  shorter  than  any  other  line-segment  drawn  from 
that  point  to  the  same  line. 

116.  DISTANCE.     The    length    of    the    perpendicular 
from  a  point  to  a  line  is  the  distance  from  the  point  to 
the  line. 

117.  ALTITUDE  OF  A  TRIANGLE.     The  distance  from 
the  vertex  of  a  triangle  to  the  base  (produced  if  neces- 
sary) is  the  altitude  of  the  triangle.      The  word  "alti- 
tude ' '  is  often  used  to  indicate  the  line  itself  as  well  as 


RECTILINEAR  FIGURES  47 

its  length.  Since  any  one  of  the  sides  of  a  triangle  may 
be  considered  as  the  base,  every  triangle  has  any  one  of 
three  possible  altitudes. 

118.  MEDIAN  OF  A  TRIANGLE.  The  sect  connecting  the 
vertex  of  a  triangle  with  the  mid-point  of  the  opposite 
side  is  a  median  of  the  triangle.  There  are  in  every  tri- 
angle three  medians. 

1.  The  bisectors  of  the  angles  at  the  base  of  an  isosceles  tri- 
angle together  with  the  base  form  another  isosceles  triangle. 

2.  If  a  line  is  drawn  from  any  point  in  the  bisector  of  an 
angle  parallel  to  one  side  of  the  angle  and  is  extended  to  meet 
the  other  side,  an  isosceles  triangle  is  formed. 

3.  A  ABC   is   a  right   triangle  with   the   vertex   of   the   right 
angle  at  C.     Draw  CD  to  AB  so  that  L  ACT)  =  L  A.     Prove  that 
D  is  the  mid-point  of  AB. 

SUG.    Prove  AD  -  CD  -  BD. 

4.  Prop.  XXII  is  the  converse  of  what  proposition? 

5.  What  relation  does  the  mid-point  of  the  hypothenuse  of  a 
right  triangle  sustain  to  the  three  vertices  of  the  triangle?     See 
Ex.  3. 

6.  To  measure  a  distance  between  two   points,  A  and  B,  both 
of  which  are  inaccessible.  A 

SUG.     1.       Lay    off   a     convenient 
line   CD    and    measure    CD,   Lo, 
L  7>i,  and  Z  n. 

2.  On  CD  as  base  construct  &ACD  and  A  BCD. 

3.  This  fixes   two  points   corresponding  to  A  and  B, 
giving  the  required  distance.    It  is  to  be  noted  that  in  these  problems 
the  method  requires  the  construction  of  triangles  which,  if  the 
distances  are  large,  may  be  impracticable. 

7.  Prove    that    AC  >  BC  —  AB    in    figure   §  112.     Why    wns 
the  proof  as  given  sufficient  to  cover  this  problem? 

8.  Using  the  dividers  only  show  that  BC  <    AB  +  AC. 

9.  Can  a   triangle  be   constructed  with  sides   of  7   ft.,   3   ft., 
and  13  ft.?     Of  7  ft.,  5  ft.,  and  12  ft.?     Of  3  ft.,  8  ft.,  and   10 
ft.?     Use  the  dividers  in  the  construction. 


48  PLANE  GEOMETRY 

PROPOSITION  XXV. 

119.  THEOREM.    //  two  sides  of  a  triangle  are 
unequal,  the  angles  opposite  these  sides  are  un~ 
eaual  and  the  angle  opposite  the  greater  side  is 
the  greater. 

Given  A  ABC  with  AB  >  AC. 

To  Prove  Z  C>  Z  B. 

Proof.  If  Z  C  is  not  greater  than  Z  B  then 
Z  C=/.B  or  L  C<£B. 

Suppose  Z  C  =  Z  B.  What  conclusion  follows  as  to 
sides  AB  and  AC°!  Auth. 

What  then  of  the  supposition? 

Suppose  Z  C  <  Z  B.  What  conclusion  as  to  the  sides 
ABandACt  Auth. 

What  then  of  this  supposition  ? 

What  possibility  remains? 

Therefore— 

Discussion.  Observe  that  one  of  three  relations  must  exist  be- 
tween Z  B  and  Z  C  and  the  true  relation  was  determined  by  prov- 
ing the  impossibility  of  the  two  others.  The  argument  does  not 
differ  from  other  indirect  arguments  that  have  been  used  except 
that  there  are  three  possible  situations  to  be  considered  instead 
of  two,  as  heretofore. 
Prove  Prop.  XXV  by  a  direct  proof. 

Sue.     On  AB  lay  off  AM  =  AC  and  draw  CM. 

/.AMO/.B.     Why? 

.'./.ACM  >  Z£. 

Z  ACS  >  /.ACM. 

.'.£ACB>/.B.     Why?  B^ ^c 

PROPOSITION  XXVI. 

120.  THEOREM.     //  sects  are  drawn  from,  the 
same  point  in  a  perpendicular  to  a  line  and  meet 


RECTILINEAR  FIGURES 


49 


the  line  at  unequal  distances  from  the  foot  of  the 
perpendicular,  the  sects  are  unequal  and  the 
more  remote  is  the  greater. 


M 

with 


0 


Given    OD  1 EB 
ED  >  DP. 
To  Prove  OE  >  OF. 

SUG.     1.     Lay  off 
tween  E  and  D.    Why? 
2.     Draw  OM. 
3. 


D 

any 


-B 


point    in    OD    and 


M  will  lie  be- 


Then  OM  =  OF.    Why  ? 
L  1  is  a  rt.  angle.     •'•  L  2  is  acute. 


Why? 


4. 
5. 
6. 
7. 

8. 


Why  ? 

•  Z  4  is  acute.    Why? 
•L3>L4.    Why? 
OE>OM.     Why? 
0#  >  0^.    Why  ? 
Therefore— 

PROPOSITION  XXVII. 

121.  THEOREM.  The  shortest  line  from  a  point 
to  a  straight  line  is  the  perpendicular  from  the 
point  to  the  line. 


Let  PA  represent  the  shortest  line  from  P  to  line  AB. 
To  Prove  PA  1  AB, 


50  PLANE  GEOMETRY 

Proof.  If  PA  is  not  perpendicular  to  AB  then  some 
other  line  through  P  is  perpendicular  to  AB.  But  by 
§115  this  new  line  would  be  shorter  than  PA,  hence 
contrary  to  hypothesis. 

Therefore— 

PROPOSITION  XXVIII. 

122.  THEOREM.  Two  unequal  line  segments 
drawn  from  the  same  point  in  the  perpendicular 
to  a  given  line  meet  the  line  at  unequal  distances 
from  the  foot  of  the  perpendicular,  the  longer  seg- 
ment meeting  it  at  the  greater  distance. 


V  JB  D 

Given  AB  1 MN  and  AC  >  AD. 
To  Prove  BC  >  BD. 

SUG.     1.     If  CB  is  not  greater  than  BD  then 
CB  =  BD  or  else  CB  <  BD. 

2.  Suppose    CB  =  BD)    how    does    AC 
compare  with  AD?     Why? 

3.  Suppose    CB<BD,    how    does    AC 
compare  with  ADt    Why? 

4.  What  then  is  the  true   relation   of 
CB  to  BD? 

Therefore— 

Of  what  theorem  is  this  the  converse  ? 

1.  A  line  drawn  from  one  end  of  the  base  of  an  isosceles 
triangle  perpendicular  to  the  opposite  side  makes  with  the  base 
an  angle  equal  to  one-half  of  the  vertex  angle. 

Suppose  the  vertex  angle  is  obtuse. 


RECTILINEAR  FIGURES  51 

PROPOSITION  XXIX. 

123.  THEOREM.  //  two  triangles  have  two 
sides  of  the  one  equal  respectively  to  two  sides  of 
the  other  and  the  included  angles  unequal,  the 
remaining  sides  are  unequal  and  that  side  is  the 
greater  which  is  opposite  the  greater  of  the  in- 
cluded angles. 

Given  A  ABC  and  A  A'B'-C'  with  AB  =  A'Bf, 
=  A'C',  and  L  A  >  L  A'. 

To  Prove  BC  >  B'C'. 


SUG.  1.  Place  AA'B'C'  upon  A  ABC  so 
that  A'B'  coincides  with  AB.  Then  since 
L  A  >  LA'  side  A'C'  will  lie  between  AB  and 
AC.  The  point  C'  may  fall  within  A  ABC,  on 
line  BC,  or  beyond  line  BC. 

The    student    should    construct    the    figures    for    the    first    and 
second  cases  lettering  them  like  the  first  figure. 

2.  Bisect    L  CACf   and  extend  the  bi- 
sector to  meet  BC  at  0. 

3.  Compare  A  AOC  with  A  AOC'  and 
OC  with  OC'.    Auth's. 

4.  Compare  EC'  with  BO  +  OC' . 

5.  Compare  BO  +  OC'  with  BO  +  OC 
or  its  equal  BC. 

Therefore — 


52  PLANE  GEOMETRY 

PROPOSITION  XXX. 

124.  THEOKEM.     //  two  triangles  have  two  sides 
of  the  one  equal  respectively  to  two  sides  of  the 
other  and  the  third  sides  unequal,  the  included 
angles  are  unequal  and  that  angle  is  the  greater 
which  is  opposite  the  greater  side. 

Given    A  ABC    and     AA'B'C',    with    AB  =  A'B', 
AC  =  A'C',  and  BOB'C'. 
To  Prove  L  A  >  L  A'. 

SUG.     1.     What    three    possible    relations    are 
there  for  A.  A  and  A'  ? 

2.  Test  each  of  these  as  was  done  in 
Prop.  XXV. 

3.  Which  is  the  only  one  not  contra- 
dicted by  the  hypothesis? 

Therefore— 

125.  QUADRILATERAL.     A  portion  of  a  plane  bounded 
by  four  sects  is  a  plane  quadrilateral. 

126.  QUADRILATERALS    CLASSIFIED.       A  quadrilateral 
is  a  parallelogram  if  its  opposite  sides  are  parallel,  a 
trapezoid  if  but  one  pair  of  opposite  sides  is  parallel, 
a  trapezium  if  no  two  sides  are  parallel.     A  trapezoid 
the  non-parallel  sides  of  which  are  equal  is  isosceles, 

127.  PARALLELOGRAMS  CLASSIFIED.      A  parallelogram 
is  a  rectangle  if  all  its  angles  are  right  angles,  a  rhom- 
boid if  all  its  angles  are  oblique,  a  square  if  it  is  an  equi- 
lateral rectangle,  a  rhombus  if  it  is  an  equilateral  rhomboid. 

With  dividers  (or  scale)  and  protractor  determine  the 
character  of  each  of  the  following  figures. 


RECTILINEAR  FIGURES  53 

128.  DIAGONAL.     A  line-segment  connecting  any  two 
non-adjacent  vertices  of  a  figure  is  a  diagonal. 

129.  BASE    OF    A    QUADRILATERAL.     The    side    upon 
which  a  quadrilateral  is  assumed  to  stand  is  its  base.    In 
case  of  the  trapezoid  or  parallelogram,  one  of  the  parallel 
sides  is  always  considered  as  the  base.    The  opposite  side 
is  the  upper  base. 

130.  ALTITUDE.    The  perpendicular  distance  between 
the  bases  of  a  parallelogram  or  a  trapezoid  is  its  alti- 
tude.    The  word  "altitude"  is  often  used  to  indicate 
the  line  itself  as  well  as  its  length. 

•  131.  DIAMETER.  The  sect  which  connects  the  mid- 
points of  two  opposite  sides  of  a  parallelogram  or  the 
mid  points  of  the  non-parallel  sides  of  a  trapezoid  is  a 
diameter. 

Construct  each  of  the  figures  above  defined  and  .draw  for  each 
the  altitude  and  indicate  the  bases  and  diameters. 

PKOPOSITION  XXXI. 

132.     THEOREM.     The  opposite  sides  of  a  par- 
allelogram are  equal. 


c 

Given  U  ABCD. 
To  Prove  AB  ~  DC  and  AD  =  CB. 

SUG.     1.     Draw  a  diagonal,  as  AB,  and  com- 
pare A  ABC  with  A  CDA.     Auth. 

2.     Compare  AD  with  CB  AB  with  CD. 
Therefore — 


54  PLANE  GEOMETRY 

133.  COR.     1.     A   diagonal  divides  a  parallelogram 
into  two  congruent  triangles. 

134.  COB.     2.     Segments  of  parallel  lines  intercepted 
betiveen  parallel  lines  are  equal. 

135.  COR.     3.     Perpendicular  segments  between  par- 
allel lines  are  equal. 

136.  COR.     4.     The    perpendicular    segment    inter- 
cepted between  the  base  of  a  triangle  and  a  line  through 
the  vertex  parallel  to  the  base  is  equal  to  the  perpen- 
dicular from  the  vertex  to  the  base. 

137.  DISTANCE.       The  length   of  the   perpendicular 
segment  intercepted  by  two  parallel  lines  is  the  distance 
between  them. 

PROPOSITION  XXXII. 

138.  THEOREM.    The  opposite  angles  of  a  par- 
allelogram are  equal  and  any  two  consecutive  an- 
gles are  supplementary. 

See  §  97  and  §  102. 

PROPOSITION  XXXIII. 

139.  THEOREM.     The  diagonals  of  a  parallelo- 
gram bisect  each  other. 

The  construction  of  the  figure  is  left  to  the  student. 

Given  O  ABC D  with  diagonals  AC  and  BD  inter- 
secting at  0. 
To  Prove  AO  =  OC  and  BO  =  OD. 

SUG.     Select  appropriate  A  and  compare  them. 
Therefore— 

PROPOSITION  XXXIV. 

140.  THEOREM.     //  a  quadrilateral  has  two  of 
its  opposite  sides  equal  and  parallel,  it  is  a  par- 
allelogram. 


RECTILINEAR  FIGURES  55 

The  construction  of  the  figure  is  left  to  the  student. 
Given  a   quadrilateral   ABCD,   with   AB  =  DC  and 
AB  II  DC. 
To  Prove  ABCD  a  /Z7. 

SUG.     1.     What  is  a  parallelogram? 

2.  How  much  of  the  definition  is  given 
in  the  hypothesis  and  what  remains  to  be  proved  ? 

3.  Draw    a   diagonal   as   AC,  compare 
L  BAG  with  L  ACD  and  then  compare  A  ABC 
with  A  ACD.    Complete  the  demonstration. 

Therefore— 

141.  A  careful  distinction  must  be  drawn  between 
statements  which  are  definitions,  i.e.,  which  are  neces- 
sary and  at  the  same  time  sufficient  to  characterize  the 
thing  defined  as  distinct  from  all  other  things,  and 
statements  which  point  out  certain  properties  of  the 
thing  in  question  without  being  in  themselves  complete 
enough  to  make  this  distinction.  For  example — since 
the  hypothesis  of  Prop.  XXXIV  is  sufficient  to  charac- 
terize the  quadrilateral  as  a  parallelogram  according  to 
the  definition  in  §  126,  it  might  be  taken  as  the  defini- 
tion while  Props.  XXXI  and  XXXII  could  not  be  so 
used  for  the  properties  there  indicated  are  possessed  by 
other  figures  than  parallelograms. 

1.  If  one  angle  of  a  parallelogram  is  a  right  angle,  the  figure 
is  a  rectangle. 

l\  The  sum  of  the  angles  of  a  quadrilateral  equals  four  right 
angles. 

3.  If  the   diagonals  .of  a   quadrilateral  bisect  each  other,   the 
figure  is  a  parallelogram.     This  bears  what  relation  to  §139? 

4.  If    the    opposite    angles    of    a    quadrilateral    are   equal    the 
figure  is  a  parallelogram. 

Sue.     Compare  the  sum   of  two  consecutive  angles  with 
four  right  angles. 


56  PLANE  GEOMETRY 

PROPOSITION  XXXV. 

142.  THEOREM.     A  quadrilateral  the  opposite 
sides  of  which  are  equal  is  a  parallelogram. 

The  construction  of  the  figure  is  left  to  the  student. 
Given  O  ABCD  with  AB  =  DC  and  AD  =  BC. 
To  Prove  ABCD  a  O. 

SUG.     1.     Draw  the  diagonal  AC. 

2.  The  conditions  of  §  126  or  §  140  may 
be  used  to  identify  the  parallelogram. 

3.  What  part  of  either  is  in  the  hy- 
pothesis of  this  theorem  ? 

4.  What  remains  to  be  proved?     Com- 
plete the  demonstration. 

Therefore— 

In  the  light  of  §  141  what  relation  does  the  statement 
of  this  theorem  bear  to  the  parallelogram? 

PROPOSITION  XXXVI. 

143.  THEOREM.     Two     parallelograms     which 
have  two  sides  and  the  included  angle  of  the  one 
equal  to  two  sides  and'  the  included  angle  of  the 
other  respectively  are  congruent. 


D  C       D' 

Given £17 AC  and  A'C',  with  AB  =  A'B', 

A'D',and  L  A= /.  A'. 
To  Prove  O  AC  =  O  A'C'. 

SUG.  1.  Place  OAO  upon  OA'G"  so  that 

AB  coincides  with  A'B'. 


RECTILINEAR  FIGURES  57 

2.  What     direction     does     AD     take? 
Why?     Where  does  D  fall?     Why? 

3.  What  direction   does  DC  take   and 
why  ? 

4.  Where  does  point  C  fall  and  why? 

5.  Complete  the  demonstration. 
Therefore— 

PROPOSITION  XXXVII. 

144.  THEOREM.     The  diagonals  of  the  rhombus 
and  of  the  square  bisect  the  angles. 

The  proof  is  left  to  the  student. 

PROPOSITION  XXXVIII. 

145.  THEOREM.     The  diagonals  of  the  rhombus 
and  of  the  square  are  perpendicular  to  each  other. 


Given  rhombus  ABCD   (or  a  square)  with  diagonals 
AC  and  BD. 

To  Prove  AC  LED. 

SUG.     1.     In    A  ADD    and     A  AOB    compare 
L  AOD  and  AOB. 

2.  Complete  the  demonstration. 

3.  Note  that  no  reference  is  made  to 
the  character  of  the  angles  A,  B,  C,  D.  What  con- 
clusion can  be  drawn  as  to  the  application  of  the 
demonstration  to  the  square  as  well  ^s  the  rhom- 
bus? 

Therefore — 


58  PLANE  GEOMETRY 

PROPOSITION  XXXIX. 

146.  THEOREM.     A    diameter    of    a    parallelo- 
gram   divides   it   into    two    congruent   parallelo- 
grams. 

The  proof  is  left  to  the  student. 

147.  COR.     1.     A    diameter    of    a   parallelogram    is 
parallel  to  the  corresponding  base. 

148.  COR.     2.     The  diameters  of  a  parallelogram  bi- 
sect each  other. 

149.  COR.     3.     A  line  which  is  parallel  to  the  base 
of  a  parallelogram  and  which  bisects  one  side  bisects  the 
other  side  also. 

PROPOSITION  XL. 

150.  THEOREM.     A  line  which  bisects  one  side 
of  a  triangle  and  is  parallel  to  the  base  bisects 
the  other  side  and  equals  half  the  base. 


Given  A  ABC  with  MN  I!  CB  and  M  mid-point  of  AC. 
To  Prove  AN  =  NB  and  2  MN  =  CB. 
SUG.     1.     Draw  MD  \\  AB. 

2.  Compare    A  AMN    and    MCD ;  MD 
and  AN  ;  MD  and  NB ;  AN  and  NB.    Auth. 

3.  Compare  MN  and  CD;  MN  and  DB; 
MN  and  CB.   Auth. 

Therefore— 

].        The  angle   formed  by  the  bisectors   of  the   angles  at  the 
base  of  an  isosceles  triangle  is  equal  to  an  exterior  angle  at  the 
of  the  triangle. 


RECTILINEAR  FIGrRES  50 

151.  COR.     The  line  connecting  the  mid-points  of  the 
sides  of  a  triangle  is  parallel  to  the  base  and  equal  to 
one  half  the  base. 

SUG.     1.    Work  out  an  indirect  demonstration. 
2.     Work  '  out    a    direct    demonstration 
from  the  above  figure,  noting  that  MN  II  CB  if 
MNBD  is  a  O  or  if  Z  AMN  —  L  MCD.    §  150. 

PKOPOSITION  XLI. 

152.  THEOREM.     //  a  line  bisects  one  of  the 
non-parallel  sides  of  a  trapezoid  and  is  parallel 
to  the  base,  it  bisects  the  other  side  alsof  and 
equals  half  the  sum  of  the  bases. 

SUG.     Draw  a  diagonal  of  the  trapezoid. 
The  proof  is  left  to  the  student. 

153.  COR.     The  diameter  of  a  trapezoid  is  parallel 
to  the  bases  and  equal  to  half  their  sum. 

SUG.  1.  Through  one  end  of  the  diameter 
draw  a  line  parallel  to  the  bases  and  use  the  in- 
direct method.  Or 

2.  Through  one  end  of  the  diameter 
draw  a  line  parallel  to  the  opposite  side  and  use 
the  direct  method. 

1.  If  the  adjacent  sides  of  a  parallelogram  are  not  equal  the 
diagonals  are  not  perpendicular  to  each  other. 

2.  If    the    diagonals    of    a    parallelogram    intersect    at    right 
angles  the  figure  is  a  rhombus  or  a  square. 

3.  If  the  diagonals  of  a  parallelogram  are  equal  the  figure  is 
a  rectangle. 

4.  The    opposite   sides    of    a   window    frame    are   equal.      Be- 
fore putting  on  stay  laths  the  carpenter  "  squares  it."     At  how 
many  corners  should  he  test  it? 


60 


PLANE  GEOMETRY 


PROPOSITION  XLII. 

154.  THEOKEM.  //  parallel  lines  intercept 
equal  segments  on  one  transversal,  they  intercept 
equal  segments  on  all  transversals. 


\A 


\. 

I 

\ 

/• 

:  v 

L 

Given  the  parallel  lines  a,  &,  c,  c?,  etc.,  with  transver- 
sals e  and  e'  such  that  AB  =  BC  =  CD. 
To  Prove  A'£'  =  5'G"  —  C'D'. 

SUG.     1.     e  and  e'  may  or  may  not  be  paral- 
lel.   What  figures  are  formed  in  each  case? 

2.  In   case   e   and  e'   are   not   parallel 
draw  lines  through  A,  B,  C,  etc.,  parallel  to  e'  and 
terminating  on  the  next  following  of  the  parallels. 

3.  Complete  the  proof. 
Therefore— 

1.  Having  laid  out   a  plat  or   four  sided  frame  so  that  the 
opposite  sides  are  equal,  how  can  one  tell  whether  or  not  it  is  a 
rectangle  without  applying  a  " square"  or  right  angle  to  one  of 
the  angles? 

2.  A     draftsman     wishing    to 
draw  parallel  lines  uses  a  T-square. 
Explain    the    principle    of  its   use. 
How  may  parallel  lines  be  drawn 
tvith  a  carpenter's  square? 

3.  Can   a   rectangle    and  a   rhombus   have 
sides   respectively  equal?     Construct  a  rectangle 

of  wood,  one  nail   at  each  vertex.         Can  it  be     distorted  into  a 
rhombus? 


RECTILINEAR  FIGURES  61 

4.        Explain  the  principle  of  the  parallel  rulers. 


3.  By  experiment  test  which  has  the  greater  enclosed  area, 
a  rectangle  or  a  rhomboid,  the  sides  of  the  two  being  respectively 
equal.  Squared  paper  can  be  used. 

6.  Construct  a  parallelogram  having  given  two  adjacent  sides 
and  the  included  angle,  using  dividers  and  straight  edge.     How 
could  it  be  done  with  a  protractor  and  straight  edge! 

7.  Draw   a  trapezium   connecting  the  mid-points  of  its   adja- 
cent sides.     What  kind  of  a  quadrilateral  is  formed? 

Sue.     Draw  a  diagonal  of  the  given  figure. 

8.  Prove     that    the    sects    that     connect    the     mid-points    of 
opposite  sides  of  a  trapezium  bisect  each  other. 

The  following  problems  are  intended  for  use  with  pupils  who 
have  had  or  who  are  taking  elementary  Physics. 

9.  Two    balls    of   the    same   size    and    rolling   with    the    same 
speed   at    right    angles    to    each   other    on   a    level    surface    strike 
another   ball   which   is   at    rest.     What    direction   will    the    latter 
ball  take?     If  one  ball  has  twice  the  force  of  the  other,  what  di- 
rection will  the  third  ball  t;ike? 

NOTE.  The  force. which  would  give  to  the  third  ball  the  same 
motion  as  is  given  to  it  by  the  two  balls  is  called  the  resujtant 
force  or  resultant  of  the  forces  due  to  the  two  balls  independently. 

10.  Draw    a    parallelogram    to    represent    the    forces    and    di- 
rections   of    the   three    balls    in    the    first    part    of    Ex.    9.      This 
parallelogram  is  called  the  parallelogram  of  forces. 

31.  Draw  the  parallelogram  of  forces  for  the  second  part  of 
problem  9,  representing  a  foot  by  %  inch. 

12.  Suppose  the  balls  in   Ex.   9   with   equal   force   strike   the 
third  ball  at  an  angle  of  45°.     Draw  the  parallelogram  of  forces 
and   a   line  representing   the  resultant   force. 

13.  Two  forces,  one  of  5  Ibs.  and  one  of  8  Ibs.,  strike  a  body 
at   an  angle   of   60°.     Draw   the  resultant.      If   they   meet   at   an 
angle  of   100°,   draw  the  resultant.     If  one  is  a   force  of   7   Ibs. 
and  the  other  21  Ibs.   and  they  meet  at  an  angle  of   100°,  draw 
the  resultant. 


62 


PLANE  GEOMETRY 


155.  POLYGONS  CLASSIFIED.     Beginning  with  the  tri- 
angle   and    proceeding    in    order    to    polygons   of    four 
sides,  five  sides,  etc.,  the  plane  polygons  are  the  trian- 
gle,  quadrilateral,   pentagon,   hexagon,  heptagon,  octo- 
gonf  etc.     A  polygon  of  n  sides  is  called  an  n-gon. 

156.  A   polygon   is   equilateral   if   all   its   sides    are 
equal,  equiangular  if  all  its  angles  are  equal,  and  regu- 
lar if  it  is  both  equilateral  and  equiangular. 

Two  polygons  are  mutually  equilateral  if  the  sides  of 
the  one  are  equal  respectively  to  the  sides  of  the  other, 
mutually  equiangular  if  the  angles  of  one  are  equal  re- 
spectively to  the  angles  of  the  other. 

The  student  should  prove  that  two  figures  both 
mutually  equilateral  and  mutually  equiangular  are  con- 
gruent. 

157.  A  polygon  is  convex  if  each  of  its  angles  is  less 
than  a  straight  angle  and  concave  if  one  or  more  of  its 
angles  are  reflex.     In  this  hook  the  term  polygon  will 
mean  a  convex  polygon. 


EQUIANGULAR    EQUILATERAL   REGULAR       CONCAVE 

158.     COR.     In  any  polygon  the  number  of  vertices 
is  the  same  as  the  number  of  sides. 

1.  If  ABC  is  a  right  triangle  with  the  right  angle  at  C  and 
CD  is  drawn  to  the  hypotenuse  so  that  LACD  —  LE,     then  CD  1 
AB.    Prove  L  DCB  =  LA. 

2.  If   two   parallel   straight   lines   are    cut   by  a   transversal, 
the  bisectors  of  the  interior  angles  form  a  rectangle. 

3.  What  relative  positions   of  parallels  and   transversals  will 
form  a  square? 


RECTILINEAR  FIGURES  63 

PROPOSITION  XLIII. 

159.  THEOREM.  The  sum  of  the  angles  of  a 
polygon  is  equal  to  (n  —  2)  straight  angles, 
where  n  represents  the  number  of  sides. 


Given  polygon  ABCD  . . . ,  having  n  sides. 
To  Prove  Z^  +  Z5+ZCf  +  etc.  =  (M  -  2)  straight 
angles. 

SUG.     1.     Draw    all    possible    diagonals    from 
some  one  vertex  as  A. 

2.  Then   (n  —  2).  triangles  are  formed. 
Explain  why. 

3.  How    many    straight    angles    in    the 
sum  of  all  the  interior  angles  in  all  these  trian- 
gles? 

4.  Complete  the  proof. 
Therefore— 

160.  COR.     I.     The  sum  of  the  angles  of  a  polygon 
is  equal  to  (2n  —  4)  rt.  angles,  or  2n  rt.  L  — 4rt.  A. 

161.  COR.     II.     Each  angle  of  an  equiangular  poly- 

7  4     2-n  -  4          , 
gon  is  equal  to  -      —  rt.  A. 
n 

1.  What    is   the   sum    of   the   angles    of    a    quadrilateral?      A 
pentagon?    A  heptagon?     A  decagon? 

2.  What  is  the  size  of  each  angle  of  a  regular  hexagon?     A 
regular  octagon?     A   regular   decagon? 


64  PLANE  GEOMETRY 

3.  How  many  sides  has  a  regular  polygon  if  each  angle  is  156°  ? 

4.  Four  of  the  angles  of  a  pentagon  are  100°,  170°,  95°,  and 
115°   respectively.     Find  the  fifth  angle. 

SUG.      Form   an    equation   from    Prop.  XLIII  and  solve 
algebraically. 

PROPOSITION  XLIV. 

162.  THEOREM.  //  one  exterior  angle  be 
formed  at  each  vertex  of  a  polygon,  the  sum  of 
these  exterior  angles  is  equal  to  four  right  angles. 


c 
Given  polygon  MC,  having  n  sides  with  one  exterior 

angle  formed  at  each  vertex,  as   La'    formed  at  ver- 
tex A  with  interior  Z  a. 
To  Prove   L  a'  +  L  b'  +  Z  c'  +  . . .  =  4  rt.  A. 

SUG.  1.  The  polygon  has  n  vertices.  What  is 
the  sum  of  the  exterior  and  the  interior  angle  at 
each  vertex? 

2.  What  is  the  sum  for  all  vertices? 

3.  What  is  the  sum  of  the  interior  an- 
gles? 

4.  Complete  the  proof. 
Therefore— 

163.     COB.     The  sum  of  all  the  exterior  angles  of  any 
polygon  is  equal  to  eight  straight  angles. 

1.  How   many   degrees   in    each   exterior   angle    of    a    regular 
pentagon?     Of  a  regular  octogon? 

2.  How  many  sides  has  the  polygon,  the  sum  of  the  exterior 
angles  of  which  is  equal  to  the  sum  of  the  interior  angles .' 

3.  An    exterior    angle    of    a    regular    polygon    is    60°.      How 
many  sides  has  the  polygon?     45°? 


RECTILINEAR  FIGURES  65 

4j  An  exterior  angle  of  a  regular  polygon  is  £  of  the 
adjacent  angle.  Find  the  number  of  sides  of  the  polygon. 

5.,  Each  angle  of  a  regular  polygon  is  £  of  a  straight  angle. 
Find  the  number  of  the  sides  of  the  polygon. 

164.  Locus.     To  locate  a  point  definitely  in  a  plane 
two  conditions  limiting  its  position  must  be  given.     If 
but  one  condition  is  given  the  point  is  limited  to  one  or 
more  lines.    For  example,  if  all  that  is  known  of  a  point 
is  that  it  is  four  inches  from  a  given  straight  line,  the 
point  is  not  definitely  fixed  in  position  but  lies  on  either 
of  two  lines  and  may  be  any  point  on  either.    The  lines 
in  question  are  parallel  to  the  given  line,  lying  the  one 
on  one  side  and  the  other  on  the  other  side  of  the  given 
line,  at  four  inches  distance,  as  will  later  be  established. 

The  locus  of  a  point  is  defined  as  the  line  or  group  of 
lines  to  which  a  point  is  limited,  any  point  of  which 
satisfies  the  given  conditions.  The  locus  of  a  point  is 
both  inclusive  and  exclusive..  It  includes  all  the  points 
which  satisfy  the  given  condition  and  excludes  all  points 
which  do  not. 

165.  Locus  GENERATED.    The  locus  of  a  point  may  be 
regarded  as  the  path  of  a  point  which  moves  according 
to  a  given  law  or  condition.     For  example,  the  law  in 
the  illustration  used  in  the  preceding  section  is  that  the 
point  must  keep  at  the  distance  of  four  inches  from  the 
given  line.    The  expressions,  locus  of  a  point  and  locus 
of  points  are  both  used. 

Discuss  without  formal  proof  the  following  loci: 

1.  What  is  the  locus  of  a  ship  at  sea  20°  N.  Latitude  f 

2.  A  man  lives  one  block  west  of  the  north  and  south  street 
which  passes  the  school  house.     What  is  the  locus  of  his  house? 

3»  A  man  is  ^  mile  from  the  street  or  road  in  front  of  the 
school.  What  is  his  locus? 


66  PLANE  GEOMETRY 

4.  An  article  was  lost  ten  feet  west  of  the  outer  edge  of  a 
straight  side  walk.    Where  should  one  look  for  it,  i.  e.  what  is  its 
locus? 

5.  In  Ex.  4  change  the  words  "west  of"  to  "from."     What 
is  the  locus? 

6.  What   is    the   locus    of    the   center   of    the   headlight   of   a 
locomotive  when  in  motion? 

7.  A  farm  is  in  range  seven,  what  is  its  locus?     In  township 
fifteen,  what  is  its  locus? 

8.  What  is  the  locus  of  the  tip  of  the  pendulum  of  a  clock? 

9.  What  is  the  locus  of  the  hub  of  a  wheel  on  a  moving  auto? 

166.  To  VERIFY  Locus  CONDITIONS.     To  prove  that  a 
given  line  (or  lines)  is  the  locus  of  points  satisfying  a 
given  condition  it  is  necessary  to  prove  two  things;  first 
— any  point  in  the  line  satisfies  the  given  condition  and 
second — no  point  outside  the  line  satisfies  the  given  con- 
dition.    For  an  example,  see  §  167. 

PROPOSITION  XLV. 

167.  THEOREM.     The  locus   of  points  equally 
distant  from  the  extremities  of  a  given  line  seg- 
ment is   the  perpendicular  bisector   of  the  line 
segment.  M  x 


N 


Given  line  segment  AB,  with  MN  the  perpendicular 
bisector  of  AB. 

To  Prove  MN  the  locus  of  points  equidistant  from  A 
and  B. 

SUG.     1.     See  Prop.  V. 

2.  Let  X'  be  a  point  not  on  MN.    Com- 
pare X'A  and  X'B.    Auth. 

3.  Apply  definition  of  locus. 
Therefore— 


RECTILINEAR  FIGURES  67 

168.  COR.     The  locus  of  points  equidistant  from  two 
given  points  is  the  perpendicular  bisector  of  the  line 
joining  them. 

PROPOSITION  XL VI. 

169.  THEOREM.     The  locus  of  points  equidis- 
tant from  the  sides  of  an  angle  is  the  bisector  of 
the  angle. 

See  examples  1  and  2  on  p.  37. 

170.  COR.     The  locus  of  points  equidistant  from  two 
intersecting  lines  is  the  pair  of  lines  that  bisect  the  an- 
gles formed  by  the  given  lines. 

What  relation  do  these  locus  lines  bear  to  each  other  ? 

PROPOSITION  XL VII. 

171.  THEOREM.    The  locus  of  points  at  a  given 
distance  from  a  given  line  is  the  pair  of  lines  par- 
allel to  the  given  line  and  at  the  given  distance 
from  it. 

SUG.  1.  Through  any  point  in  the  given  line 
draw  a  perpendicular,  on  this  line  locate  the 
two  points  which  are  at  the  required  distance 
from  the  given  line  and  through  these  points 
draw  lines  parallel  to  the  given  line. 

2.  What  two  facts  must  be  established 
in  order  to  prove  these  two  constructed  lines  to  be 
the  desired  locus? 

Therefore— 

1.     If  the  opposite  angles  of  a  parallelogram  are  bisected  by 
the  diagonals,  the  figure  is   equilateral. 
Sue.     Draw  one  diagonal. 


68  PLANE  GEOMETRY 

i 
PROPOSITION  XL VIII. 

172.  THEOREM.    The   locus    of   points   equally 
distant  from  two  parallel  lines  is  the  line  parallel 
to  them  and  midway  between  them. 

SUG.     1.     Construct  a  segment  perpendicular 
to  and  intercepted  by  them. 

2.  Construct     the     perpendicular     bi- 
sector of  this  segment. 

3.  Complete  the  proof. 
Therefore — 

173.  USE  OF  Loci.     A  common  method  of  locating  a 
point  in  a  plane  is  to  establish  two  loci  of  the  point,  the 
intersection  of  which   is  the   required  point.     To   use 
this  method  of  attack  successfully  the  student  must  bear 
in  mind  that  two  statements  are  made  when  a  line,  as 
AB,  is  said  to  be  the  locus  of  an  unknown  point  X,  viz : 
any  point  in  AB  can  be  X  and  no  point  outside  of  AB 
can  be  X,  i.e.  X  must,  be  in  AB. 

1.  Find   point  X  if  it  is  to  be  in  a   given  line  and  equally 
distant  from  two  points. 

GIVEN  line  CD  and  the  points  A  and  B  (which  may  lie,  one  or 
both,  on  CD).  To  find  point  X  in  CD  and  equally  distant  from 
A  and  B. 

SUG.    1.      What    is   the  locus   of   points   equally   distant 
from  A  and  5? 

2.  Where  then  must  X  lie? 

3.  What    position    of    the    points    A    and    B    with 
respect  to  CD  would  make  the  problem  impossible? 

2.  Find  point  X  if  it  is  in  a  given  line  and  equally  distant 
from  two  intersecting  lines. 

SUG.   1.      What  is  the  locus  of  points  equally  distant 
from    two    intersecting    lines? 

2.  Complete    the    demonstration. 

3.  Is  this  problem  ever  impossible?     What  is  the 
condition? 


RECTILINEAR  FIGURES  69 

4.     What  relative  position  of  the  given  lines  would 
admit  of  an  unlimited  number  of  positions  for  point  XI 

3.  Find  point  X  if  it  is  equally  distant  from  two  intersecting 
lines   and  equally   distant   from   the   extremities  of   a  given  line- 
segment.     Is  there  any  arrangement  of  the  given  lines  which  will 
make  this  problem  impossible? 

4.  Find  point  X  if  equally  distant  from  the  sides  of  a  given 
angle  and  equally  distant  from  two  parallel  lines. 

5.  Find  point  X  if  equally  distant  from  two  points  and  equally 
distant  from  two  parallel  lines. 

6.  Find  point  X  if  it  lies  in   one  given  line  and  is  equally 
distant    from    another    given    line.      Are    there    conditions    which 
make  the  problem  impossible? 

7.  Find  X  if  it  is  equally  distant  from  two  intersecting  lines 
and  also  a  given  distance  from  a  third  line. 

In  general,  how  many  such  points  are  there!  Whatsis  the 
least  possible  number?  What  is  the  greatest  possible  number  f 

174.  CONCURRENT  LINES.     Two  or  more  lines  having 
one  point  in  common  are  concurrent  lines. 

The  construction  of  loci  is  a  useful  method  of  proving 
certain  lines  concurrent.  Name  the  concurrent  lines  in 
the  above  exercises. 

1.  The  perpendicular  bisectors  of  the  sides  of  a  triangle  are 
concurrent. 

SUG.  1.  Two  of  the  bisectors  as  DE 
and  FG  musF  mfersect  as~aT~#T— W-hy  ? 
What  loci  are  here  involved? 

2.  OA  =  OC  and  OC  =  OB.  Why?    C^  F  %B 

3.  Therefore  0  is  equidistant  from  B  and  A.    Why? 

4.  In  what  third  line  must  0  then  lie?     Why? 

2.  The  bisectors  of  the  angles  of  a  triangle  are  concurrent. 

SUG.    1.     Two    of    the    bisectors    must   intersect.    Why? 
What  loci  are  involved? 

2.  Prove  their  intersection  to  be  in  the  third  bisector j 

175.  SUMMARY  OF  BOOK  I. 

State  all  the  authorities  by  which 

(1)     two  triangles  may  be  proved  congruent. 


70  PLANE  GEOMETRY 

(2)  two  line  segments  may  be  proved  equal. 

(3)  two  lines  may  be  proved  parallel. 

(4)  two  lines  may  be  proved  perpendicular  to 
each  other. 

(5)  two  angles  may  be  proved  unequal. 

(6)  two  line  segments  may  be  proved  unequal. 

(7)  a  quadrilateral  may  be  proved  a  parallel- 
ogram. 

(8)  a  locus  has  been  established. 

These  authorities  should  be  carefully  arranged,  writ- 
ten out  in  a  note  book,  and  memorized.  They  are  the 
"tools"  to  be  used  in  future  constructions  and  demon- 
strations. Geometry  is  not  half  learned  if  the  student 
is  unable  to  recall  the  various  hypotheses  which  are 
available  for  deducing  any  desired  conclusion. 

1.  BD  bisects  L  ABC  and  EF  LBD  at 
B.       Prove     L  FBA  =  L  CBE.       Prove     the 
theorem  when  EF  cuts  BD  at  some  point 
other  than  B. 

2.  If  two   straight  line  segments  bisect 
each    other    at    right    angles    any    point    in 
either    is    equidistant    from   the   extremities 
of  the  other. 

3.  By  how  much  does  the  supplement  of  an  acute  angle  exceed 
the  complement  of  the  same  angle? 

4.  If  one  of  two   supplementary  adjacent   angles   is  bisected, 
the  perpendicular  to  this  bisector  at  the  vertex  bisects  the  other  angle. 

5.  If  the  bisectors  of  two  adjacent  angles  are  perpendicular 
to  each  other  the  angles  are  supplementary.     Compare  with  Ex.  4. 

6.  The  perimeter  of  a  triangle  is  leis  than  twice  the  sum  of 
the  medians. 

7.  The  bisector  of  an  angle  of  a  triangle  and  the  bisectors  of 
the  exterior  angles  of  the  two  other  vertices  meet  in  a  point  which 
is  equidistant  from  the  sides  of  the  triangle.     Use  loci. 

8.  If   a   line   intersects   the   sides   of  an  isosceles   triangle   at 
equal  distances  from  the  vertex;  it  is  parallel  to  the  base. 


RECTILINEAR  FIGURES  71 

9.  The   line  joining  the   feet  of  the   perpendiculars   from  the 
extremities  of   the   base   of  an   isosceles   triangle   to    the   sides  is 
parallel  to  the  base. 

10.  Two   angles   having   their  sides   respectively   parallel,    one 
pair   of  parallel   sides   having  the   same   direction   and   the   other 
pair  having  opposite  directions,  are  supplementary. 

11.  Two  triangles  having  two  angles  and  a  side  opposite  one 
of  them  in  the  first  equal  respectively  to  two   angles  and  a  corre- 
sponding side  in  the  other  are  congruent. 

12.  Two  isosceles  triangles  having  equal  bases  and  equal  vertex 
angles  are  congruent. 

13.  Two    triangles   having    two    sides    and   an    angle    opposite 
one  of  them  in  the  one  triangle  equal  to  two  sides  and  an  angle 
opposite  one  of  them  respectively  in   the   other  triangle  may  or 
may  not  be  congruent.  A  D 


GIVEN  A  ABC  and 
A  DEF  or  £  D'E'F', 
with  AB=DF  =  D'F', 
AC  =  DE  =  D'E',  L  B  =  L  F  =  L  F'. 

To  PROVE  A  ABC  =  A  DEF   but  not   congruent  to  A  D'E'F'. 

It  is  to  be  noted  that  of  the  two  triangles  DEF  and  D'E'F'  one 
is  acute  and  one  is  obtuse,  and  each  have  the  required  parts. 
Obviously  A  ABC  is  congruent  to  but  one.  The  student  should 
give  detailed  proof  for  the  figures  as  drawn. 

14.  The  sum  of  the  exterior  angles  at  the  base  of  a  triangle 
is  equal  to  two  right  angles  plus  the  vertex  angle. 

15.  In   &ABC,   LC  is  twice  the   sum   of   L  A    and   ZB   and 
LE  is  twice  LA.     Find  A,  B,  C. 

16.  The  exterior  angles  formed  by  producing  the  sides  of  an 
isosceles  triangle  beyond  the  base  are  equal. 

17.  Given   a  square  ABCD.     Draw   the   diagonal   CD  and   on 
CD  lay  of   CE  =?CB.     Draw   EF  LCD  at   E  and  extend  to   DB 
as  at  F.    Prove  DE  =  EF  =  BF. 

18.  A  straight  line  segment  intercepted  between  parallels  is 
bisected  and  another  straight  line  is  drawn  through  the  point  of 
bisection.     Prove  that  the  segment  of  this  line  intercepted  between 
the  parallels  is  bisected. 


72 


PLANE  GEOMETRY 


19.  Find  the  sum  of  the  interior  angles  of  a  concave  polygon 
of  13  sides. 

20.  A    segment    is    intercepted 
between     two     parallels     and     the 
adjacent     angles    formed    by    the 
segjnent   and   one   of  the   parallels 
are 'bisected.     Prove  that  the  seg- 
ments   on    the    other    parallel    are 
equal. 

21.  ABC   is   an   isosceles    triangle.     DE 
and    DF    are    parallel    to    AC   and    AB    re- 
spectively.    Prove   the  perimeter  of  AFDE 
equal  to  AB  +  AC. 

22.  If  the  legs  of  a  trapezoid  are  equal, 
they  make  equal  angles  with  the  parallel  sides. 

23.  Prove  the  converse  of  Ex.  22. 

24.  Prove  that  the  sum  of  the  angles  at  the 
vertices   of  the    conventional   five   pointed    star 
is  equal  to  two   right   angles.     Use   the  figure 
as  drawn. 

25.  Connect  the  vertices  of  the  star  and  give 
a  different  demonstration. 

26.  How  many  right  angles  in  the  sum  of  the  vertex  angles  of 
a  six  pointed  star?     Of  an  eight  pointed  star? 

27.  If  the  vertex  angle  of  an  isosceles  triangle  is  one-half  as 
great  as  an  angle  at  the  base,  the  bisector  of  a  base  angle  di- 
vides the  given  triangle  into  two  isosceles  triangles. 

28.  How    many    equiangular    triangles    can    have    a    common 
vertex?        How     many     rectangles?       How     many     equiangular 
pentagons?      Hexagons?     Equiangular   figures    of   a  still   greater 
number  of  sides? 

29.  Which  of  the  above  figures  will  exactly  fill  up  the  space 
about  the  common  vertex? 

30.  Which  of  the  above  figures  can  be  used  for  a  patch  work 
quilt  or  a  mosaic  design? 


CHAPTER  II. 


CIRCLES. 

176.  CIRCLE.     A  closed  curve  in  a  plane  all  points 
of  which  are  the  same  distance  from  a  fixed  point  in  the 
plane  is  a  circle.      The  fixed  point  is  the  center  of  the 
circle.     The  circle  completely  encloses  a  portion  of  the 
plane.    Thus  a  circle  is  the  locus  of  points  in  a  plane  at 
a  given  distance  from  a  fixed  point  in  the  plane. 

It  has  been  more  or  less  customary  for  writers  on 
geometry  to  define  the  circle  as  the  portion  of  the  plane 
enclosed  by  the  curve  and  to  call  the  curve  the  circum- 
ference of  the  circle.  The  above  use  of  the  term  is  more 
convenient  and  accords  more  perfectly  with  its  general 
use. 

177.  LENGTH  OP  A  CIRCLE.     The  length  of  the  curve 
is  called  the  length  of  the  circle. 

178.  RADIUS.     A  straight  sect  joining  the  center  and 
any  point  of  the  circle  is  a  radius. 

179.  CHORD.     A  straight  sect  terminated  at  both  ends 
by  the  circle  is  a  chord. 

180.  DIAMETER.    A  chord  which  passes  through  the 
center  is  a  diameter. 

181.  ARC.     Any  portion  of  a  circle  is  an  arc. 

An  arc  which  is  half  of  a  circle  is  a  semicircle.  An 
arc  less  than  a  semicircle  is  a  minor  arc  and  one  which 
is  greater  than  a  semicircle  is  a  major  arc. 

182.  CENTRAL  ANGLE.     An  angle  the  vertex  of  which 


74  PLANE  GEOMETRY 

is  the  center  of  a  circle  and  the  sides  of  which  are  radii 
is  a  central  angle. 

183.  SECANT.     A  straight  line  that  intersects  a  cir- 
cle twice  is  a  secant. 

A  secant  is  a  chord  produced.  .x v 

OA   and    OB    are    radii,    AB   is  a       /      o__—-V 
diameter,   CD  is  a  chord  and  ED  is      AT  cy 

a  secant.  °^^_ ^^ 

184.  SUBTENDS.     A    chord    subtends    the    two    arcs 
which  have  the  same  extremities  as  itself.     Unless  other- 
wise stated  the  arc  subtended  by  a  chord  is  understood 
to  be  the  minor  arc. 

185.  PRELIMINARY  THEOREMS  AND  ASSUMPTIONS. 

1.  All  radii  of  a  circle  are  equal. 

2.  The  diameter  of  a  circle  is  twice  the  radius. 

3.  All  diameters  of  a  circle  are  equal. 

4.  The  distance  from  any  point  in  the  plane  to  the 
center  of  a  circle  is  greater  than,  less  than,  or  equal  to 
a  radius  according  as  the  point  is  outside  the  circle,  on 
the  circle,  or  within  the  circle. 

5.  If  the  radii  or  diameters  of  two  circles  are  equal 
the  circles  are  equal.     Since  all  circles  are  of  the  same 
shape  this  implies  congruence  as  well. 

6.  If  two  circles  are  equal,  the  radii  and  the  diam- 
eters of  the  two  are  equal. 

7.  An  unlimited  straight  line  that  lies  partly  within 
a  circle  cuts  the  circle  in  two  points. 

8.  If  the  end  points  of  two  minor  arcs  or  two  major 
arcs  on  the  same  or  on  equal  circles  can  be  made  to  coin- 
cide the  arcs  can  be  made  to  coincide. 

9.  If  two  arcs  of  two  circles  are  equal  the  circles  are 
equal. 


CIRCLES  75 

186.  POSTULATE.     A  circle  may  be  constructed  about 
any  given  point  in  the  plane  as  center  ivith  any  given 
sect  as  radius. 

The  actual  construction  of  such  a  circle  is  done  by  aid 
of  the  dividers. 

187.  COR.     A   circle  may  be   constructed   upon  any 
given  sect  as  diameter. 

SUG.     What  must  be  done  to  the  given  diameter 
to  find  the  center  of  the  required  circle  ? 

188.  PROBLEM.     Given  an  arc  of  a  circle,  to  find  the 
center. 

SUG.     If    from   the    given   arc 
MC  two  loci  of  the  center  can  be 
determined,    the    center    can    be     M 
found. 

2.  Draw  two  chords   as  AB  and  AC. 
Where  must  the  center  lie  with  respect  to  A  and 
Bf     With  respect  to  A  and  Cf 

3.  Construct   the   loci  of  points  which 
satisfy  these  respective  conditions.     What  can  be 
said  of  the  intersection  of  these  locij 

4.  A  circle  about  this  point  0  as  a  cen- 
ter with  radius  OA  will  pass  through  points  A, 
B,   C  and  therefore   by   §  185  (8)    will   embrace 
the  given  arc. 

Therefore,  point  0  is  the  required  center. 

189.  PROBLEM.     To    draw    a    circle    through    three 
points  not  in  the  same  straight  line. 

SUG.     Make  the  construction  from  §  188,  with 
A,  B,  C  as  given  points  in  required  O. 

190.  COR.  I.    A  circle  cannot  be  drawn  through  three 
points  which  are  in  a  straight  line. 


76  PLANE  GEOMETRY 

i 

191.  COB.  II.     Only     one     circle     can     be     drawn 
through  three  points  not  in  a  straight  line. 

SUG.  It  was  seen  in  §  189  that  at  least  one 
such  circle  can  be  drawn.  From  §  185  (8)  it  fol- 
lows that  there  cannot  be  more  than  one.  Why? 
Or,  take  a  point  X  as  any  other  center.  Can  X  be 
equally  distant  from  A  and  B  f  from  A  and  C  f 

1.  How  many  circles  can  be  drawn  through  two  points? 

2.  Two  circles  can  intersect  in  two  points  at  most. 

3.  Given  arc  M,  find  the  center. 

PROPOSITION  I. 

192.  THEOREM.    In  the  same  circle,  or  in  equal 
circles,  equal  central  angles  intercept  equal  arcs. 


Given  circle  0  equal  to  circle  0'  and 
L  AOB  =  LA'O'E'. 
To  Prove  arc  AB  =  arc  A'B'. 

Proof.  SUG.  1.  Place  OO  on  OO'  with  0  on  0'. 
Then  the  two  circles  will  coincide,  having  a  com- 
mon center  and  equal  radii. 

2.  Rotate  OO  about  0'  until  point  A 
falls  on  A'. 

3.  Then  since  L  AOB  =  L  A'O'B'  the 
radii  OB  and  O'B'  will  coincide  and  B  fall  on  B' . 

4.  Hence  the  arcs  AB  and  A'B'  coin- 
cide.    §185(8). 

Therefore— 

1.     How  can  a  carpenter  test  the  accuracy   of  his  "square" 
without  using  a  previously  made  right  angle? 


CIRCLES  77 

PROPOSITION  II. 

193.  THEOREM.    In  the  same  circle,  or  in  equal 
circles,  equal  arcs  subtend  equal  central  angles. 

Given  O  0  and  O  0'  with  arc  AB  =  arc  A'B' '.     (Use 
fig.  Prop.  I.) 
To  Prove  Z  AOB  =  Z  A'O'B', 

SUG.     Place  the  circles  together  as  in  Prop.  I  and 
show  that  the  two  angles  can  be  made  to  coincide. 

Therefore— 

PROPOSITION  III. 

194.  THEOREM.     //,  in  the  same  or  in  equal 
circles,  two  central  angles  are  unequal,  the  greater 
angle  intercepts  the  greater  arc. 


Given  O  C  equal  to  O  C"  and  Z  ACB  <  L  A'C'B'. 
To  Prove  arc  AB  <  arc  A'B'. 

Proof.    SUG.  1.     Place  O  C  upon  O  C'  as  in  Prop.  I 
and  rotate  it  until  CA  coincide  with  C'A'. 

2.  Where  will  CB  fall  with  reference 
to  ZA'C"£'?     Why,? 

3.  Where  will  B  fall  with  reference  to 
thearcA'5'?     Why? 

4.  Compare  arcs  A'B    and  A'B' ;  arcs 
AB  and  A'B'. 

Therefore— 

1.     A  diameter  is  greater  than  any  other  chord. 

Sue.  Draw  any  chord  not  a  diameter  and  draw  radii 
to  its  extremities.  Compare  the  chord  with  the  sum  of  the 
radii. 


78  PLANE  GEOMETRY 

PROPOSITION  IV. 

195.  THEOREM.    //'  two  arcs,  in  the  same  or  in 
equal  circles,  are  unequal,  the  greater  arc  sub- 
tends the  greater  central  angle. 

Given  O  C  equal  to  O  6"   and  arc  AB  <  arc  A'B'. 
(Use  fig.  of  Prop.  III.) 
To  Prove  Z  ACE  <  L  A'C'B'. 

Proof.     SUG.     Place  O  C  upon  OC"  as  in  Prop.  Ill 
and  show  that  Z  ACS  equals  a  part  of  /.A'C'B'. 
Therefore — 

PROPOSITION  V. 

196.  THEOREM.    In  the  same  circle,  or  in  equal 
circles,  equal  chords  subtend  equal  arcs. 


Given  O  F  and  OF'  equal  with  chord  AB  =  chord 
A'B'. 

To  Prove  arc  AB  =  arc  A'  £', 

Proof.    SUG.  1.     Draw  the  radii  FA,  FB,  F'A'  F'B'. 

2.     The  arcs  are  equal  provided 
LF-=LF'.    Why? 

3.     Prove    these    angles    equal    and 
complete  the  demonstration. 

Therefore— 

1.  Napoleon  and  his  engineer  in  exploring  came  to  a  river. 
Napoleon  asked  its  width.  The  engineer  sighted  from  the  rim  of 
his  cap  to  the  opposite  side,  swung  upon  his  heel,  and  sighted  to 
a  point  on  the  land,  then  paced  to  the  point  and  said  "Ten  rods, 
Sire."  Upon  what  proposition  did  his  computation  depend? 


CIRCLES  79 

PROPOSITION  VI. 

197.  THEOREM.    In  the  same  circle,  or  in  equal 
circles,  chords  which  subtend  equal  arcs  are  equal. 

Given  two  equal  ®,  F  and  F'  with  arc  AB  =  arc  A'B' . 
(Use  fig.  of  Prop.  V.) 
To  Prove  chord  AB  =  chord  A'B'. 
Proof.    SUG.     Compare  the  central  angles  F  and  F' 
and  complete  the  demonstration. 

Therefore— 

PROPOSITION  VII. 

198.  THEOREM.     In    the    same    circle,    or    in 
equal  circles,  two  chords  which  subtend  unequal 
minor  arcs  are  unequal .  and  the  greater  chord 
subtends  the  greater  arc. 


Given  equal  ©  E  and  E'  with  arc  AB  <  arc  A'B'. 

To  Prove  chord  AB  <  chord  A'B'. 

Proof.     SUG.     1.     Draw  radii  EA,  EB,  E'A',  E'B'. 

2.  Compare  1  E  and  E'.     Auth.? 

3.  Compare  chords  AB  and  A'B'. 
Therefore— 

1.  If  the  mid-points  of  the  three  sides  of  a  triangle  bo  joined 
by  straight  lines,  the  triangle  is  divided  into  four  congruent  tri- 
angles. 

2.  If  the  mid-points  of  two  opposite  sides  of  a  quadrilateral 
be  joined  to  the  mid-points   of  the   diagonals,   the   joining  lines 
form  a  parallelogram.    As  one  particular  case  let  the  quadrilateral 
be  a  parallelogram.     Ts  this  case  an  exception? 


80  PLANE  GEOMETRY 

PROPOSITION  VIII. 

199.  THEOREM.    In  the  same  circle,  or  in  equal 
circles,  two  arcs  which  subtend  unequal  chords 
are   unequal  and   the  greater  arc  subtends   the 
greater  chord. 

Given   equal   ©   E  and  E'   with   chord   AB  <  chord 
A'B'.     (Use  fig.  of  Prop.  VII.) 
To  Prove  arc  AB  <  arc  A'B'. 

Proof.     SUG.     1.     Compare   A.  E  and  E' .     By  what 
theorem  ? 

2.     Complete  the  demonstration. 
Therefore— 

PROPOSITION  IX. 

200.  THEOREM.    In  the  same  circle,  or  in  equal 
circles,  equal  chords  are  equally  distant  from  the 
center. 


Griven  equal   ©    C   and   C'   with   chord   AB  —  chord 
A'B'. 

To  Prove  AB  and  A'B'  equally  distant  from  C  and 
C'  respectively. 

Proof.     SUG.     1.     Draw  CM  LAB  and  C'M'LA'B'. 
Why?    Also  draw  CB  and  C'B.'     Why? 

2.     It  is  necessary  to  prove 
CM=C'M'.    Why? 

3.     Complete  the  proof. 
Therefore — 


CIRCLES  81 

PROPOSITION  X. 

201.  THEOREM.    In  the  same  circle,  or  in  equal 
circles,  chords  which  are  equally  distant  from  the 
center  are  equal. 

Given  ©  C  and  C'  with  chords  AB  and  A'R'  such 
that  their  distances  CM  and  G'M'  from  the  respective 
centers  are  equal.  (Use  fig.  of  Prop.  IX.) 

To  Prove  chord  AB  =  chord  A'B'. 

Proof.     SUG.     Compare    MB    and  M'B'.      Complete 
the  demonstration. 

Therefore— 

PROPOSITION  XI. 

202.  THEOREM.    In  the  same  circle,  or  in  equal 
circles,    unequal    chords    are    unequally    distant 
from  the  center  and  the  shorter  chord  is  at  the 
greater  distance. 


Given  O  C  with  chord 
AB  <  chord  ED,  CM  and  CN  be- 
ing the  respective  distances  of  the 
chords  from  the  center. 

To  Prove  CM  >  CN. 

^ — : 

D 

Proof.     SUG.     1.  Draw  chord  FD  =  AB  and 
CG  1 FD.    Connect  O  and  N. 

2.  Compare  Z  x  with  Z  y.     Auth. 

3.  Hence  Z  u  <  Z  v.     Why  ? 

4.  '''CG>CN.    Why? 

5.  •'.  CM>CN.     Why? 
Therefore — 


PLANE  GEOMETRY 

PROPOSITION  XII. 

203.  THEOREM.    In  the  same  circle,  or  in  equal 
circles,  chords  which  are  unequally  distant  from 
the  center  are  unequal  and  that  chord  ivhich  is  at 
the  greater  distance  is  the  shorter. 

Given  O  C,  CM  and  CN  being  the  respective  distances 
of  two  chords  AB  and  ED  from  the  center.    Also 
CN  <  CM.     (Use  fig.  of  Prop.  XL) 
To  Prove  AB  <  ED. 

Proof.     SUG.     1.     \7hat  three  possibilities  exist  as  to 
the  relative  sizes  of  AB  and  EDI 

2,     Assume   each  in  turn  to   be   true. 
Which  is  the  only  one  which  does  not  by  a  former 
theorem   lead   to    a    conclusion    contradictory   to 
the  hypothesis? 
Therefore— 
Prove  prop.  §  203  by  direct  method  (use  fig.  §  202). 

PROPOSITION  XIII. 

204.  THEOREM.     A  radius  perpendicular  to  a 
chord  bisects  the  chord. 


Given  O  C  with  chord  AB  and  radius  CN  1  AB  at  .V. 

To  Prove  AN  =  NB. 

Proof.    SUG.    1.     Draw  radii  CA  and  CB.    Why  ? 

2.  Compare      A  CNA     with     A  CNB. 
Auth. 

3.  Compare  AN  and  NB.     Auth. 
Therefore— 


CIRCLES  83 

205.  COR.     1.     A  radius  perpendicular   to  a  chord 
bisects  the  arc  subtended  by  the  chord. 

SUG.     Compare  the  two  arcs  by  means  of  the 
subtended  central  angles. 

206.  COR.     2.     A   radius   ivhich    bisects    a   chord  is 
perpendicular  to  the  chord. 

207.  COR.     3.     The     perpendicular     bisector    of     a 
chord  of  a  circle  bisects  the  arc  subtended  by  the  chord. 

208.  COR.     4.     The  perpendicular  bisector  of  a  chord 
of  a  circle  passes  through  the  center  of  the  circle. 

1.  If  the  straight  line  connecting  the  mid-points  of  two  chords 
of  a  circle  passes  through  the  center,  the  two  chords  are  parallel. 

2.  The  radius   drawn  to   the   mid-point  of  an   arc  is  the  per- 
pendicular bisector  cf  the  subtended  chord. 

209.  TANGENT  LINE.     A  line  that  touches 
a  circle  at  one  point  only  and  does  not  cut 
the  circle  is  a  tangent  line. 

210.  POINT  OF  TANGENCY.     The  point  which  is  com- 
mon to  the  circle  and  a  tangent  line  is  the  point  of 
tangency. 

211.  TANGENT  CIRCLE.     If  two  circles  have  but  one 
point  in  common  they  are  tangent  circles. 

If  one  circle  is  within  the  other  they 
are  tangent  internally.  If  they  are  with- 
out each  other  they  are  tangent  exter- 
nally. 

1.  In  a  right  triangle  with  acute  angles  of  30°   and  60°  re- 
spectively, one  side  is  one-half  the  hypothenuse. 

2.  If  the  hypothenuse  of  a  right  triangle  is  equal  to  twice 
one  of  the  sides  the  acute  angles  are  30°  and  60°  respectively. 

3.  The   mid-point    of    the    hypothenuse    is    equidistant    from 
the  three  vertices. 


84  PLANE  GEOMETRY 

PROPOSITION  XIV. 

212.  THEOREM.  The  straight  sect  joining  the 
centers  of  two  tangent  circles  passes  through  the 
point  of  tangency. 


Given  two  circles  C  and  C',  tangent  at  M. 
To  Prove  line  CC'  to  pass  through  M. 
Proof.     CASE  I.     Internal  tangency. 

SUG.  1.  Assume  that  CC'  does  not  pass 
through  M. 

Join  C  to  M  and  C'  to  M.  Extend  CC'  to 
meet  O  C  at  N  and  O  C'  at  N' . 

The  pupil  will  note  that  the  figure  is  distorted 
for  the  sake  of  the  argument.  The  points  C' 
C'  are  not  the  true  centers. 

2.  Compare  CM  and  CN.     Auth? 

3.  CN>CN'.     "Why? 

4.  CN'  —  CC'  +  C'N'. 

5.  C'N'  =  C'M.    Why? 

6.  ••"•  <73f  >  CC'  +  C'M.     Is  it  possible  ? 

7.  What  of  the  assumption  in  step  1? 

CASE  II.     External  tangency. 

1.  Make    the    same    assumption    as    in 

case  I. 

2.  CM  =  CN  and  CN  <  CN'.    Why  ? 

3.  CW'  +  CW  =  CC'. 

4.  CN+C'N'<CC'.     Why? 


CIRCLES  85 

5.  •'•  CM  +  C'M  <  CC'.     Why?  Is  this 
possible  ? 

6.  What  of  the  assumption  in  step  1? 
Therefore— 

213.  COR.     //  two  circles  are  tangent  a  line  tangent 
to  one  at  their  point  of  contact  is  tangent  to  the  other. 

PROPOSITION  XV. 

214.  THEOREM.     A    straight   line   perpendicu- 
lar to  a  radius  at  its  outer  extremity  is  tangent 

to  the  circle.  A     M    ^ 


Given  0  C  with  line  AB  1  CA  at  its  outer  extremity  A. 
To  Prove  AB  tangent  to  O  C. 

Proof.     SUG.  1.     What  must  he  known  to  prove  AB 
tangent  ? 

2.  How  much   of  this   is   included   in 
the  hypothesis? 

3.  What  remains  to  be  proved? 

4.  Let  M  represent  any  point  on  AB 
other  than  A  and  draw  CM. 

5.  Compare  CA  and  CM  as  to  length. 

6.  Where   then    is    point    M   with    re- 
spect to  O  (7? 

Therefore— 

215.     COR.     I.     A  line  which  is  tangent  to  a  circle  is 
perpendicular  to  the  radius  at  the  point  of  contact. 

SUG.     Prove  that  the  radius  is  the  shortest  line 
from  the  center  to  the  tangent. 


86  PLANE  GEOMETRY 

216.  COR.     II.     The  perpendicular  to  a  tangent  at 
the  point  of  contact  passes  through   the  center  of  the 
circle. 

SUG.     Indirect  proof. 

217.  COB.     III.     The   perpendicular   dropped   from 
the  center  of  a  circle  to  a  tangent  meets  it  at  the  point 
of  tangency. 

218.  COB.     IV.     At  any  point  on  a  circle  one  and 
only  one  tangent  can  l)e  drawn. 

PROPOSITION  XVI. 

219.  THEOREM.    Arcs   of  a   circle   intercepted 
by  parallel  lines  are  equal. 

Given  O  C,  with  two  parallel  chords  AB  and  DE  in- 
tercepting arcs  EB  and  AD. 
To  Prove  arc  EB  =  arc  AD. 
Proof.     SUG.  1.      Drop  a  1  from  C 

to  DE,  extending  it  to  meet  the 

circle  as  at  M.    Why  ? 

2.  How  does  CM  lie  with  reference  to 
AB1 

3.  How  does   CM  effect  the   arcs  sub- 
tended by  the  two  chords? 

4.  Complete  the  demonstration. 

5.  Assume    one    of    the    two    parallel 
lines  to  be  tangent  to  the  circle  and  complete  the 
demonstration. 

6.  Assume    both   lines   to   be   tangents 
and  complete  the  demonstration. 

Therefore — 


CIRCLES  87 

220.  SEGMENT  OF  A  CIRCLE.  A  portion  of  the  plane 
enclosed  by  an  arc  and  its  subtended  chord  is  a  seg- 
ment of  the  circle. 

ABC  is  a  minor  segment  and   CD  A   is  a 
major     segment.     By     segment     is     usually 


meant  the  minor  segment  and  it  is  indicated    V  / 

by  its  arc  alone. 

221.  SECTOR  OF  A  CIRCLE.     A  portion  of  the  plane 
enclosed   by  two    radii   and   the   intercepted    arc    is   a 
sector  of  the  circle.     DOA  is  a  sector. 

222.  CIRCUMSCRIBED  POLYGON.    A  poly- 
gon is  circumscribed  about  a  circle  if  all  of 
its  sides  are  tangent  to  the  circle,  as  poly- 
gon ABCDE. 

When  a  polygon  is  circumscribed  about  a  circle, 
the  circle  is  inscribed  in  the  polygon. 

223.  INSCRIBED  POLYGON.      A    polygon 
is  inscribed  in  a  circle  if  each  of  its  sides 
is  a  chord  of  the  circle,  as  A'B^'J)'. 

When  a  polygon  is  inscribed  in   a   circle,   the  circle  is  en  cum- 
scribed  about  the  polygon. 

224.  INSCRIBED  ANGLE.     An  angle  is  inscribed  in  a 
circle  when  its  vertex  is  on  the  circle  and  its  sides  are 
chords. 

225.  ANGLE  INSCRIBED  IN  A  SEGMENT.     An  angle  is 
inscribed  in  a  segment  when  it  is  subtended  by  the  chord 
of  the  segment  and  its  vertex  is  on  the  arc  of  the  segment. 

226.  Whenever  it  is  established  that  a  certain  relation 
of  points  and  lines  is  possible,  the  rigor  of  a  demon- 
stration is  not  impaired  by  using  a  representation  of 
that  relation  without  performing  the  actual  construe- 


88  PLANE  GEOMETRY 

t 

tion.  This  has  been  done  directly  in  previous  demon- 
strations and  indirectly  by  the  use  of  instruments,  such 
as  the  protractor,  the  construction  of  which  required 
the  results  of  theorems  at  that  time  not  proven.  This 
has  been  done  in  order  that  the  pupil  may  at  as  early 
a  stage  as  possible  learn  to  make  careful  constructions 
and  more  fully  visualize  the  relations  under  consideration. 

Plane  geometry  deals  only  with  figures  composed  of 
straight  lines  and  circles,  the  construction  of  which  may 
usually  be  effected  by  means  of  the  straight  edge  and 
the  dividers.  The  postulates  involving  the  simple  use  of 
these  tools  are  restated  below.  Problems  of  construc- 
tion are  in  no  sense  Pure  Geometry  but  are  applications 
of  principles  demonstrated  in  Pure  Geometry. 

Since  constructions  are  based  on  previously  demon- 
strated theorems  some  problems  can  be  solved  by  several 
methods  according  as  one  or  another  theorem  is  used. 
For  example,  several  methods  for  determining  a  perpen- 
dicular have  already  been  indicated  as  well  as  for  par- 
allel lines.  In  the  following  problems  it  will  be  of 
added  interest  to  make  use  of  as  many  methods  as  pos- 
sible in  each. 

227.     POSTULATES. 

(1)  A  sect  can  be  drawn  between  any  two  points 
and  can  be  extended  to  any  length  through  either  ex- 
tremity. 

(2)  A  circle  can  be  drawn  with  any  sect  for  a  radius 
about  any  point  as  a  center. 

The  first  postulate  requires  the  straight  edge  and  the 
second  the  dividers. 

In  solving  problems  of  construction  it  is  of  advantage 
to  represent  the  construction  as  completed  in  order  that 


CIRCLES  89 

by  a  study  of  the  lines  involved  and  their  relations  to 
each  other  one  may  recall  previous  theorems  and  con- 
structions from  which  as  starting  points  the  desired 
construction  can  be  made. 

PEOPOSITION  XVII. 

228.  PROBLEM.     To    bisect    a    given    straight 
line  segment. 

Construction.  With  A 
and  B,  the  extremities,  as 
centers  and  with  equal 

radii  greater  than  1  AB  de-     A 

scribe  arcs  intersecting  as 
at  M  and  N.  Draw  MN  in- 
tersecting AB  at  0. 

Then  0  is  the  mid-point  of  AB. 

Proof.  By  the  construction  M  and  N  are  each  equi- 
distant from  A  and  B  and  hence  determine  the  perpen- 
dicular bisector  of  AB. 

NOTE.  A  little  experience  will  suggest  what  parts  of  lines  may 
be  omitted  as  unessential.  For  example,  in  the  above  construction 
all  the  circles  may  be  omitted  except  such  short  arcs  as  are  nec- 
essary to  determine  the  intersection  points  M  and  .AT". 

PEOPOSITION  XVIII. 

229.  PROBLEM.     To  erect  a  perpendicular  to  a 
given  line  at  a  given  point  on  the  line. 


M 


Given  point  M  in  line  c. 

To  Construct  a  perpendicular  to  c  at  M. 

Construction.     On  c  lay  off  MA  =  MB.    With  A  and 


90  PLANE  GEOMETRY 

B  as  centers  and  equal  radii  greater  than  MA  describe 
arcs  intersecting  at  D.  Draw  DM  which  is  the  required 
perpendicular. 

Proof.  DM  is  the  perpendicular  bisector  of  AB 
(Why?)  and  as  AB  is  a  part  of  line  c,  DM  is  perpen- 
dicular to  c. 

PROPOSITION  XIX. 

230.  PROBLEM.  From  a  given  point  to  drop  a 
perpendicular  to  a  given  line. 


/\ 


Given  the  line  c  with  a  point  M  not  on  c. 

To  Construct  a  line  through  M  perpendicular  to  c. 

Construction.  With  M  as  a  center  and  a  radius  of 
sufficient  length  describe  an  arc  to  cut  line  c  in  two 
points  as  A  and  B.  Locate  now  a  second  point,  AT,  equi- 
distant from  A  and  B.  How  can  this  be  done?  The 
line  MN  is  the  required  line. 

Proof.     Left  to  the  student. 

PROPOSITION  XX. 
231.     PROBLEM.     To  bisect  a  given 
arc. 
Given  AB,  the  arc  to  be  bisected. 

SUG.     1.     What  propositions  have  been  demon- 
strated involving  the  bisection  of  an  arc? 

2.  If  the  center  0  is  known,  complete 
the   demonstration. 

3.  If  the  center  is  not  given,  complete 
the  demonstration. 


CONSTRUCTIONS  91 

Proof  left  to  student. 

1.  Let  a  be  a  given  straight  sect  with  an  unlimited  straight 
line  intersecting  it.  Construct  a  right  triangle  with  hypothenuse  a 
and  the  vertex  of  the  right  angle  on  the  second  line. 

PROPOSITION  XXI. 

232,  PROBLEM.    To  bisect  a  given  angle. 

SUG.     1.     Construct   a   circle   with   the    given 
angle  at  the  center  and  complete  the  demonstration. 
Or         2.     What     propositions    involve     a    bi- 
sected angle  ?     Make  a  construction  from  one  or 
more  of  these. 
Proof 

PROPOSITION  XXII. 

233.  PROBLEM.     At  a  given  point  in  a  given 
line  to  Construct  an  angle  equal  to  a  given  angle 
with  the  given  line  as  one  side. 


c 
Given  L  M  and  the  point  A  on  line  AB. 

To  Construct  an  angle  at  A  equal  to  L  M  and  with 
AB  as  one  side. 

SUG.  1.  LM.  and  the  angle  to  be  constructed 
will  be  equal  if  they  are  central  angles  in  equal 
circles  and  are  subtended  by  equal  arcs.  Com- 
plete the  construction. 

2.     L  M    and    the    angle    to    be    con- 
structed will  be  equal  if  they  are  opposite  equal 
sides  in  congruent  triangles.     Complete  the  con- 
struction. 
Proof 


!)2  PLANE  GEOMETRY 

PROPOSITION  XXIII. 

234.  PROBLEM.  Through  a  given  point  without 
a  straight  line  to  construct  a  line  parallel  tothe 
given  line. 


o  _  — 


•-'c 

Given  any  point  C  not  on  line  AB. 
To  Construct  a  line  through  C  parallel  to  AB. 

SUG.     1.     Draw  through  C  any  line  cutting  AB 
as  a  transversal. 

2.  With  respect  to  AB  and  the  trans- 
versal  what   conditions  of   the   line  to  be   con- 
structed will  make  it  parallel  to  AB  ? 

3.  Complete  a  construction. 

Other  methods.    Make  constructions  by  other  authori- 
ties. 

Proof 

PROPOSITION  XXIV. 

235.     PROBLEM.     To    divide    a    sect    into    any 
number  of  eqrM  parts. 


Given  sect  a. 

To  divide  a  into  5,  or  more  generally  into  n  equal 
parts. 

SUG.     1.     From  one  extremity  of  a  draw  any 
line  oblique  to  a. 

2.     Lay  off  on  this  line  five  (or  n)  equal 
sects  of  any  convenient  length. 


CONSTRUCTIONS  93 

3.  Join    the    extremity   of   the   last   of 
these   to   the   free   extremity   of   a   and   through 
each  of  the  other  points  of  division  draw  paral- 
lels to  this  join  line  extending  them, to  meet  a. 

4.  The  five  (or  n)  divisions  thus  made 
in  a  are  equal.  .  Why  ? 

Proof 

PROPOSITION  XXV. 

236.  PROBLEM.  Given  two  angles  and  the  in- 
cluded side  of  a  triangle,  to  construct  the 
triangle. 


Given  A  A  and  B  of  a  triangle  with  A'B'  as  the  in- 
cluded side. 

To  Construct  the  triangle. 

SUG.     Represent  the  triangle  as  already  con- 
structed and  from  the  figure  decide  which  of  the 
preceding  problems  might  be  used  to  construct  it. 
Proof 

QUERY.     Can  the  line  and  angles  be  of  such  magni- 
tude as  to  make  the  construction  impossible? 

1.  Upon  a  given  base  construct  an  isosceles  triangle  in  which 
the  sum  of  the  two  other  sides  equals  a  given  line. 

2.  If  from  two  opposite  vertices  of  a  parallelogram  two  lines 
be  drawn  to  the  middle  points  of  two  opposite  sides,  the  lines  will 
trisect  the  diagonal  joining  the  other  vertices. 

The    three    medians    of    a 
triangle  meet  in  a  point. 

SUG.       CF    cuts    off    one- 
third  of  diagonal  AD.   Why? 

Find    relation    of    BE    to 
the  diagonal. 

Show    that     median     AO     lies   in    diagonal   AD. 


94  PLANE  GEOMETRY 

PROPOSITION  XXVI. 

237.  PROBLEM.  Given  two  sides  and  the  in- 
cluded angle  of  a  triangle  to  construct  the 
triangle. 


Given  two  sides,  a  and  b,  and  the  included  angle  C 
of  a  triangle. 

To  Construct  the  triangle. 

SUG.     1.     At  any  point  on  an  unlimited  line 
construct  an  angle  equal  to  C. 

2.     Complete  the  construction. 
Proof 

PROPOSITION  XXVII. 

238.     PROBLEM.     To  construct  a  triangle,  given 
the  three  sides. 


Given  a,  b,  c  as  the  three  sides  of  a  triangle  ABC. 
To  Construct  A  ABC. 

Construction.     SUG.     1.     Take    a    sect    AB    equal    to 
c.   A  and  B  are  then  two  vertices  of  the  triangle. 

2.  If    b    is    the    side    opposite 
vertex  B  what  is  the  locus  of  the  third  vertex  C? 

3.  With    respect   to   vertex   B 
where  does  C  lie? 

4.  ( -omplctc   Hie   construction. 
Proof 


CONSTRUCTIONS  95 

Discussion.  What  effect  do  the  relative  magnitudes 
of  the  given  sides  have  upon  the  possibilities  of  con- 
struction ? 

PROPOSITION  XXVIII. 

239.  PROBLEM.  To  circumscribe  a  circle  about 
a  triangle. 


Given  A  ABC. 

To  circumscribe  a  O  about  A  ABC. 

SUG.     1.     The  problem  is  to  find  the  center  of 
a  circle  which  passes  through  A,  B,  C. 

2.  Where  must  this  center  lie  with  re- 
spect to  A  and  Bf     With  respect  to  B  and  Cf 

3.  Complete     the      construction     and 
verify. 

Proof 

QUERY.     How    many    circles    can    be    circumscr  bed 
about  a  triangle?     Why? 

NOTE.     Compare  this  problem  with  §  231. 

PROPOSITION  XXIX. 

240.  PROBLEM.    To  inscribe  a  circle  in  a  given 
triangle. 

Sue.     1.     Study  Ex.  1,  P.  37,  and  write  out  the 
construction  in  full. 

2.     Prove  the  sides  of  the  triangle  to  be 
tangents  to  the  circle  thus  constructed. 

241.  REVIEW. 

State  all  the  theorems  of  Book  II  by  which  one  can 
prove 


96 


PLANE  GEOMETRY 


the    problem    as 
What  condition 


(1)  Two  lines  equal. 

(2)  Two  angles  equal. 

(3)  Two  lines  perpendicular. 

(4)  Two  lines  unequal. 

(5)  Two  angles  unequal. 

(6)  Two  arcs  equal. 

(7)  Two  arcs  unequal. 

1.  Construct    an    equilateral    triangle.      Will 
stated   admit   of  more  than  one  such   triangle? 
may  be  added  to  make  the  problem  definite  f 

2.  Construct  an  isosceles  triangle  with  the  base  one  third  of  a  side. 

3.  Construct  a  triangle  with  a  given  base,  given  base  angle, 
and  a  given  altitude. 

4.  Construct  an  angle  of  GO8  without  a  protractor. 

5.  Construct  a  triangle  with  an  angle  of  45°,  an  angle  of  60°, 
and  a  given  sect  for  the  included  side. 

6.  Construct  an  equilateral  triangle  on  a  two  inch  base.     Cir- 
cumscribe a  circle  about  it  and  inscribe  a  circle  within  it.     Show 
that  the  radii  of  the  former  is  twice  that  of  the  latter.    Show  that 
this  is  true  for  any  equilateral  triangle. 

7.  Draw    the    pattern    at    the    right 
the  square  having  a  two  inch  base. 

8.  Can  a  circle  be  circumscribed  about 
a    square?     A    rectangle?     A    rhombus? 
A  rhomboid?     Prove  your  conclusions. 

9.  Construct    a   checker-board    design 
using     a     T-square.       Construct     another 
using  a  ruler  and  a  right  triangle.     Give 
authorities. 

10.  Given  the  two  diagonals  of  a  rhombus,  construct  it.     Do 
the  same  for  a  square. 

11.  Construct  a  rectangle  having  given  one  side  and  a  diagonal. 

12.  Construct  a  rhomboid  having  given  the  two  diagonals  and 
their  included  angle. 

13.  Inscribe  a  square  in  a  given  circle;  circumscribe  one  about 
the  circle;    and  then  circumscribe  a   second  circle  about   the  sec- 
ond square.     Compare   the  length  of  the   sides  and   diagonals  of 
the  two  squares  with  the  radii  of  the  two  circles. 


CHAPTER  III. 


MEASUREMENT  AND  PROPORTION. 

242.  MAGNITUDE.     Magnitude  is  the  size,  extent,  or 
mass   of   anything   and   prompts   the    question,    "How 
much?" 

Magnitudes  in  geometry  are  lines,  angles,  areas,  solids, 
etc. 

243.  MEASURE.     To  measure  a  magnitude  is  to  find 
the  number  of  times  it  contains  a  given  unit  of  measure. 

244.  UNIT.     A  unit  of  measure  is  a  selected  magni- 
tude of  the  same  kind  as  the  magnitude  to  be  measured. 

In  everyday  experience  a  unit  of  measure  is  a  stand- 
ard set  by  statute  cr  common  consent,  e.g.  the  yard,  the 
gallon,  the  degree,  the  cubic  foot,  etc. 

245.  QUANTITY.     Quantity  is  the  result  of  measure- 
ment and  answers  the  question  "How  much?" 

The  quantities  of  geometry  are  lengths,  areas,  con- 
tents, etc. 

246.  NUMERICAL    MEASURE.     The     numerical    meas- 
ure of  a  magnitude  is  the  number  which  expresses  how 
many   times  the   unit   is   contained   in   the   magnitude 
measured. 

If  a  line  is  measured  and  found  to  be  8  feet  long,  the  line 
is  the  magnitude,  the  foot  is  the  unit,  eight  is  the  numerical  meas- 
ure, and  8  feet  is  the  quantity. 

Magnitude  is  indefinite,  quantity  is  definite.  To  express  quan- 
tity two  elements,  the  numerical  measure  and  the  unit,  are  nec- 
essary. Careful  distinction  must  be  made  between  quantity,  and 
number  which  is  the  measure  of  magnitude.  These  terms  are 


98  PLANE  GEOMETRY 

often  confused.  If  as  units  of  measure,  we  consider  the  gallon, 
the  degree,  the  square  inch,  the  cubic  foot,  then  the  expressions 
26  gallons,  5  degrees,  29  square  inches,  18  cubic  feet  are  quan- 
tities while  25,  5,  L'9,  18  are  numbers. 

247.  RATIO.  The  ratio  of  one  magnitude,  quantity, 
or  number  to  another  of  the  same  kind  is  the  number 
which  expresses  how  many  times  the  first  contains  the 
second,  or  more  generally  the  quotient  of  the  first  by  the 
second.  In  other  words,  the  ratio  of  one  magnitude, 
quantity,  or  number  to  another  is  the  numerical  meas- 
ure of  the  first,  with  respect  to  the  second  as  the  unit 
of  measure,  or,  both  being  measured  by  the  same  unit, 
the  ratio  is  the  quotient  of  the  numerical  measure  of 
the  first  by  the  numerical  measure  of  the  second. 

(a)  To  illustrate — the  ratio  of  sect  a  to  sect  &  is  the  num- 
ber of  times  a  contains  fe  as  a  unit  of  measure.     The  result  may 
be  obtained  by  laying  off  &  upon  a  as  many  times  as  possible.    If, 
however,  a  does  not  contain  &  an  integral  number  of  times  they 
may  each  be  measured  by  a  common  unit   m,  in  which  case  the 
ratio  is  the  quotient  of  the  numerical  measure  of  a  by  the  numer- 
ical measure  of  &. 

(b)  From  the  definition  of  ratio,  it  follows  that  a  ratio  can 
exist  only  between  magnitudes  or  quantities  of  the  same  kind  and 
also  that  the  ratio  is  always  an  arithmetical  number.    For  example, 
if  sect  a  contains  unit  m  8  times  and  sect  b  contains  m  4  times, 
the  ratio  of  a  to  &  is  8  -^  4  or  2.     Hence  if  c  be  the  number  of 
times  m  is  contained  in  sect  a  and  d  be  the  number  of  times  m   is 
contained  in  sect  &,    the  ratio  of  the  two  sectsis  that  of  c  to  d. 

(c)  Since    ratio    in    the    following   discussions    is    based   upon 
the  algebraic  conception  of  division  and  not  upon  the  Euclidean 
definition   (which  is  much  too  difficult  for  beginners),  the  division 
or  fractional   form   of   expression   will   be  used.      Hence  the  ratio 

of    6    ft.    to    2    ft.    will    be    written    as    <Lft-,    J>,   or   3.      The 

2  ft.      2 
statement  above  as  to  the  ratio  of  the  sects  a  and  b   will  be  short- 


MEASUREMENT  99 

248.  COMMENSURABLE.     Two  magnitudes  or  quanti- 
ties are  commensurable  if  they  each  contain  a  common 
unit  of  measure  a  whole,  or  integral,  number  of  times. 
The  ratio  of  two  commensurable  quantities  must  then 
be  an  integral  number  or  a  quotient  of  two  integral 
numbers. 

In  determining  the  common  unit  of  measure  of  two 
commensurable  magnitudes  or  quantities,  the  usual 
method  for  finding  the  greatest  common  divisor  may  be 
followed,  which  is  to  divide  the  greater  of  the  two  by 
the  less,  then  the  divisor  by  the  remainder,  this  re- 
mainder by  the  second  remainder  and  so  on,  until  a  re- 
mainder zero  is  obtained.  This  exact  divisor  is  the  de- 
sired common  unit  of  measure. 

249.  INCOMMENSURABLE  MAGNITUDES.      Two   magni- 
tudes or  quantities  which  do  not  possess  a  common  unit 
of  measure  are  incommensurable.    The  ratio  of  two  such 
magnitudes  is  neither  an  integral  number  nor  a  frac- 
tion. 

The  circle  and  its  diameter  are  incommensurable,  as  are  also 
the  diagonal  and  side  of  a  square.  A  proper  discussion  of  this 
subject  cannot  be  made  at  this  point  on  account  of  certain  diffi- 
culties necessarily  involved. 

250.  PROPORTION.     A  proportion  is  an  equality  each 
member  of  which  is  a  ratio.    The  four  numbers  or  quan- 
tities a,  b,  c,  d  are  in  proportion  if  the  ratio  of  a  to  b 
equals  the  ratio  of  c  to  d.     The  symbolic  form  of  this 

statement    is       -  — -.     The    four    numbers    or    quan- 

b      d 
tities  are  the  terms  of  the  proportion. 

a.  An  equality  of  three  or  more  equal  ratios  is  a  con- 
tinued proportion. 


100  PLANE  GEOMETRY 

1.  Arrange  the   numbers   2,   5,    20,   8  in   a  proportion   in   as 
many    ways    as    possible,    verifying    each    by    the    definition    of 
proportion. 

2.  Arrange  the  numbers  3£,  5|,  28^,  ]  7|  in  a  proportion  and 
verify  it. 

PROPOSITION  I. 

251.  THEOREM.  //  a  line  is  parallel  to  the 
base  of  a  triangle  it  divides  the  sides  into  propor- 
tional segments. 


/-A 


Given  A  ADE  with  line  BC  II  DE. 

To  Prove  ±H  =  4<}. 
BD     CE 

Proof.  SUG.  1.  To  obtain  the  ratio  of  the  sects  AB 
and  BD  they  must  be  measured.  Let  I  be  the  unit 
sect  and  suppose  it  is  contained  m  times  in  AB 

and  n  times  in  BD.  What  is  the  ratio ?  Why  ? 

BD 

2.  Draw  lines  through  the  points  of 
division  in  AB  and  BD  il  to  DE  and  extend  to 
AE.  Compare  the  segments  on  AC  and  CE. 

3.  What  is  the  ratio  of  AC  to  CE? 
Why? 

AB  AC 

4.  Compare  the  ratio  -    -   with 

BD  CE 

Therefore — 

NOTE.  In  the  above  demonstration  the  case  in  which  AB  and 
BD  have  no  common  unit  of  measure  is  not  considered.  The  con- 
clusions in  this  case,  however,  are  the  same  as  above  but  the 


PROPORTION  101 

demonstration  is  omitted  here  as  being  too  difficult  for  the  be- 
ginning student. 

252.  AN  IMPORTANT  CONSIDERATION.  Since  all  ratios 
of  magnitudes  or  quantities,  by  the  definition  in  use,  can 
be  considered  only  through  the  ratios  of  their  numerical 
measures,  all  proportions  herein  will  be  treated  as  numer- 
ical proportions.   And  since  the  ratio  of  two  magnitudes, 
as  a  and  b,  247  (b)  is  a  number  and  can  be  represented 

only  by  the   ratio  of  their   numerical   measures,  as  — • 

n 

the  terms  a  and  b  will  be  considered,  in  the  interests  of 
brevity,  to  represent,  as  well,  their  numerical  measures. 
In  general  the  names  and  notations  of  magnitudes  used 
in  the  operations  of  proportion  will  be  synonymous  with 
the  notations  for  their  numerical  measures,  and  the  treat- 
ment of  the  proportions  used  will  be  that  of  algebra. 

For  convenience  of  reference  the  propositions  of  pro- 
portion will  be  collected  and  briefly  reviewed  in  the  fol- 
lowing sections. 

1.  What  is  the  ratio  of  one  side  of  an  equilateral  triangle  to 
the  perimeter?     Of  the  perimeter  to  one  side? 

2.  What   is   the  ratio    of   a    right    angle   to    an    angle    of    an 
equilateral  triangle? 

3.  WThat  is  the  ratio  of  a  quadrant  to  a  semi  circle?     To  a 
circle? 

253.  THE  TERMS  OF  A  PROPORTION.     The  first  and 
third  terms  of  a  proportion  are  the  antecedents. 

The  second  and  fourth  terms  are  the  consequents.  The 
second  and  third  terms  are  the  means.  The  first  and 
fourth  terms  are  the  extremes. 

In   the  proportion,  -  —  —    a   and    c    are   the   ante- 

b       d 

cedents;  b  and  d  are  the  consequents;  b  and  c  are  the 
means;  a  and  d  are  the  extremes. 


102  'PLANE  GEOMETRY 

THEOREMS  OF  PROPORTION. 

254.  THEOREM  I.     The  product  of  the  means  equals 
the  product  of  the  extremes. 

SUG.     Write  the  ratios  as  fractions  and  clear 
the  equality  of  its  fractional  form. 

255.  THEOREM  II.     //  the  product  of  two  numbers 
equals   the  product  of  two  others,   the  factors  of  one 
product  may  be  made  the  means  and  the  factors  of  the 
other  product  may  be  made  the  extremes  of  a  proportion. 

Given  ab  =  cd. 

To  Prove  *  =  £ 

c       b 

SUG.     By   what   must   ab   be    divided   to   pro- 
duce —  ?     By  what  must  cd  be  divided   to   pro- 
c 

duce  -  ?      Why  are  the  resulting   ratios   equal  ? 
b 

1.  From  ab  =  cd  derive   a  proportion  in  which  a  and   b   are 
the  means. 

2.  From  ab  =  cd  make  as  many  proportions  as  possible.     Note 
in  what  respects  they  differ. 

3.  If  the  first  three  terms  of  a  proportion  are  5,  7,  15  what 
is  the  fourth  term? 

_!__§.     Find*. 

4.  Given   n  -  x 

5.,     Find  x  in  the  following  proportions: 

JL  ..-.*,     £.  --  A     11  =.  2.     I:?  _  11.1  _  jl     A  _  1? 

13  ~  18'     11  ,     13>     15  ~  27'      x    ~  39'  x  ~   12'     11  ~   x' 

6.  Make  four  different  proportions  from  the  identity 
8  X  7  =  4  X  14. 

7.  Use    theorem    I    to    determine    whether    or    not    the    ratios 

—  and  —  will  form  a  proportion.1 
12  15 

256.  THEOREM  III.       //  -  =  -»  M-™  -  =  -• 

b      d  c       d 


PROPORTION  103 

SUG.     Use  Theorems  I  and  II. 

257.  ALTERNATION.     The    interchange    of    the    two 
means  of  a  proportion  is  alternation. 

1.  State   theorem    lit    in   words    without   the   use   of   symbols 
a,  b,  c,  d. 

2.  In  the  fig.  of  §  251  take  the  proportion  by  alternation. 

0.  Construct    a    triangle    with    a    line    parallel    to    the    base. 
Measure  three  of  the  four  segments  into  which  the  two  sides  are 
cut  and  by  §  251  determine  the  fourth.     Check  the  calculation  by 
measurement. 

4.  Show  how  the  conditions  of  Ex.  3  may  be  used  to  measure 
an  inaccessible  distance,  over  a  pond  for  example. 

5.  On  the  school  grounds  drive  four  stakes  A,  B,  C,  not  in  a 
straight  line  and  D  in  the  line  AC.     Find  the  point  for  a  5th  stake 
so  as  to  measure  AD  indirectly. 

258.  THEOREM  IV.    //     -  =  -   then  -  =  -- 

1)      d  a       c 

.    a      I 

SUG.     1  —  -  —  — 
b      a 

Complete  the  proof. 

259.  INVERSION.     The  interchange  of  first  and  sec- 
ond, third  and  fourth  terms  of  a  proportion  is  inversion. 

1.  State  theorem  IV  in  words  without  the  use  of  the  symbols 
a,  b,  c,  d. 

2.  In  Ex.  5,  §  255,  take  the  proportions  by  inversion.     Which 
of  those  given  may  be  obtained  from  the  others  by  inversion? 


260.  THEOREM  V.      //   -  =  -  then  =         . 

1)      d  b          d 

SUG.     Add    1    to    each  member   of   the    given 
proportion.     Complete  the  proof. 

261.  COMPOSITION.     The  second  proportion  in  Theo- 
rem V  is  obtained  from  the  first  by  composition.     This 
process  is  sometimes  termed  addition. 

1.     State  theorem  V  in  words  without  the  use  of  the  symbols 
a,  b,  c,  d. 


104  PLANE  GEOMETRY 


2.  By  composition,  alternation,  etc.,  prove  in  the  triangle  of 

Prop.I  the  proportion  4R=EA 
BD     CE 

3.  Prove  AD  =  AE.     Sug.  §  256  and  §  260. 

AB      AC 

4.  Prove  4£  =  42. 

AC      AE 
BD      AB 


6.     Prove  =         - 

AE       CE 

262.     THEOREM  VI.    //    -  =  -,  then 


b      d  b          d 

SUG.     Subtract   1    from   each   member  of  the 
given  proportion.     Complete  the  proof. 

263.  DIVISION.     The  second  proportion  in   Theorem 
VI  is  obtained  from  the  first  by  division.    This  is  some- 
times termed  subtraction. 

7,    a      c    .,        a+b      c+d 

264.  THEOREM  VII.    //   -  =  -•  then 

b      d  a—b     c—d 

SUG.  The  second  proportion  is  said  to  be 
obtained  from  the  first  by  composition  and  divi- 
sion. It  may  be  obtained  from  the  final  propor- 
tions in  Theorems  V  and  VI.  Complete  the 
proof. 
In  the  accompanying  figure,  CD  being  parallel  to  NO,  show  that 

a_  _  b.  J>  _  d,  ,  a+c  _  b+d  .  a+c  _  b+d  .     a        _c__  .  a—  &  _  c—d  . 

c        d'    a       c'    a  b    '    c  d     '  a+b     c+d'  a  +  b      c+d' 

a  —  b  _  a+b 

c—d  ~  c+d 

2.     In   the  proportion  £L  =  _£ 

b        d  M 

a  =  12,  b  =  5,  c  =  6,  find  d. 
a  =  3,  c  =  7,  d  =  14,  find  b. 
b  =  12,  c  =  13,  d  =  6,  find  a. 
a  =  13,  b  =  15,  d  -  45,  find  c. 


PROPORTION  105 

PROPOSITION  II. 

265.  THEOREM.  //  a  line  divides  two  sides  of 
a  triangle  proportionally,  it  is  parallel  to  the 
base. 


Given        _  =     _ 
AD     AE 
To  Prove  BC  II  DE. 


Proof.     SUG.    1.     If  BC  is  not  parallel  to  DE.  let 
BM  represent  the  line  through  B  which  is. 

2.  Then  ^  =  ^.     Why? 

AD      AE 

3.  From  this  proportion  and  the  hy- 
pothesis compare  AC  and  AM. 

4.  "Where  then  must  point  M  lie  wi':h 
respect  to  Cf 

Therefore— 

1.  Given  Ex.  2,  §  264.    MN  =  1 7,  MO  =  24,  DO  =  8 ;  find  a,  ~b,  c. 

2.  Given  c  =  12,  tZ  =  6,  and  MN  =  30,  find  the  other  parts. 

266.  FOURTH  PROPORTIONAL.  The  fourth  term  of  a 
proportion  of  four  different  terms  is  a  fourth  propor- 
tional to  the  three  others  in  order. 

3.  Given  three  sects  a,  b,  c,  find  x  a  fourth  proportional  by 
construction.      Sug.      Construct    an    angle    one    side    of    which    is 
the  sum   of  the  sects  a  and   b.      On   the   second   side   at   the  ex- 
tremity of  sect  a  lay  off  sect  c.     Complete  the  construction  and 
verify  it. 

4.  Divide  a  given  sect  in  the  ratio  of  three  to  four.     Of  two 
to  three.     Of  five  to  three. 

5.  Divide  a  given  sect  a  into  two  parts  having  the  ratio  of  the 
given  sects  b  and  c. 


106  PLANE  GEOMETRY 

PROPOSITION  III. 

267.  THEOREM.  The  'bisector  of  an  angle  of  a 
triangle  divides  the  side  opposite  into  sects  which 
are  proportional  to  the  adjacent  sides. 


Given   A  ABC,  with  AM  bisecting   Z  A,  and  M  the 
point  of  division  of  CB. 


To  Prove  - 

MB     AB 

Proof.     SUG.  1.     Extend  CA  to  0  making  AO  =  AB. 
Join  0  and  B. 

2.  05  II  AM.     Why? 

3.  Note   that  AM   is  parallel   to   BO 
and  derive  the  required  proportion. 

Therefore— 

1.  If  a  line  bisects  one  side  of  a  triangle  A 
and  is  parallel  to  the  base  it  bisects  the  other                     /\ 
side  and  equals  half  the  base.                                           BX__\c 

D// AE 

Sue.     To  prove  BC  —  \  DE,  draw  a  line  through  B  par- 
allel to  AE.     Then  use  the  first  part  of  the  exercise. 
See   §  150  for  a  different  method. 

2.  If  a  line  bisects  two  sides  of  a  triangle,  it  is  parallel  to  the 
base  and  equals  one-half  the  base. 

See  §  151.     Use  another  method  here. 

3.  Divide  sect  a  of  Ex.   5  P.  105  into  tlirco  equal  parts.    Into 
four  equal  parts. 


GIVEN  AB  =  BD  and  BC  II  DE. 
PROVE  AC  -  CE  and  BC  =    £ 


PROPORTION  107 

268.     INTERNAL  DIVISION.     A  sect  is  divided  into  seg- 
ments internally   when  the   point   of 


division  lies  in  the  segment. 

AB  is  divided  internally  at  C  into  the  two  segments 
AC  and  CB.  Thus  AC  +  CB  must  equal  AB.  Ax.  9, 
§  49. 

269.  EXTERNAL  DIVISION.  A  sect  is  divided  into  seg- 
ments externally  when  the  point  of  division  lies  in  the 

extension  of  the  sect. 

A g c 

AB  is  divided  externally  at  C  into  segments  AC  and 
CB. 

In  order  that  in  this  case  also  the  sum  AC  +  CB  may 
equal  AB,  the  algebraic  idea  of.  positive  and  negative 
quantities  may  be  introduced.  In  moving  a  point  from 
A  to  B  through  the  point  of  division  C,  the  directions 
AC  and  CB  are  the  same  for  internal  division  and  op- 
posite for  external  division.  In  the  latter  case,  in  mov- 
ing from  C  to  B,  the  point  traverses  a  second  time  part 
of  segment  AC  but  with  respect  to  the  direction  of  AC, 
sect  CB  is  then  negative.  Hence  in  this  case  also  one 
may  write  AC  +  CB  =  AB.  Unless  expressly  stated  the 
notations  used  in  this  text  will  not  involve  this  use  of 
directed  lines. 

1.  Draw  a  tangent  to  a  circle  at  a  given  point  on  th'e  circle. 

2.  Two  tangents  drawn  to  a  circle  from  the  same  point  are  equal. 

3.  If  two  circles  are  concentric,  all  chords  of  the  larger  which 
are  tangent  to  the  smaller  are  equal. 

4.  The  tangents  to  a  circle  at  the  extremities  of  a  diameter 
are  parallel. 

5.  Draw  a  triangle  as  large  as  may  be  on  a  given  sheet  of 
paper  and  draw  a  line  parallel  to  the  base  and  intersecting  the 
sides.   Measure  three  parts  from  which  measurements  the   fourth 
part  may  be  found.     Check  by  measuring  this  fourth  part. 


108 


PLANE  GEOMETRY 


PROPOSITION  IV. 

270.  THEOREM.  The  bisector  of  an  external 
angle  of  a  triangle  divides  the  opposite  side  ex- 
ternally into  sects  proportional  to  the  adjacent 
sides. 


Given  A  ABC  with  L  CBE  bisected  by  a  line  divid- 
ing the  opposite  side  AC  externally  in  point  0. 

AO     AB 

To  Prove  —  =  —  • 
OC     BC 

Proof.    SUG.    1.    Draw  CM  II  BO. 


CO 


= 
MB     BC 


Therefore— 

In  the  above  figure  and  demonstration  the  inequality 
AB  >  BC  is  assumed.  Let  the  pupil  prove  the  theorem 
for  the  case  AB  L  AC  by  lettering  the  figure  the  same 
and  following  the  line  of  proof. 

1.  It  is  desired  to  measure  an  inaccessible  distance  on  a 
plane.  What  use  can  be  made  of  Proposition  §270?  What 
measurements  shall  be  taken  to  determine  AC? 


PROPORTION  100 

1.  A  line  drawn  through  the  vertex  of  a  triangle  dividing 
the  opposite  sides  into  segments  proportional  to  the  adjacent 
sides  bisects  the  angle. 

This  is  the  converse  of  what  proposition  ? 

2.  If  a  radius  of  one  circle  is  a  diameter  of  another,  the 
circles  are  tangent  and  any  line  drawn  from  the  point  of  con- 
tact to  the  outer  circle  is  bisected  by  the  inner  one. 

271.  HARMONIC  DIVISION.  A  line  is  divided  har- 
monically when  it  is  divided  internally  and  externally  in 
the  same  ratio.  *• 


M' 

1.  AB  is  divided  internally  in  the  ratio  of  6  to  4  at 
points.    AB  =  6  +  ±  =  10. 

2.  AB  is  divided  externally  at  M'  in  the  ratio  of  6 
to  4.    AB  =  6  +  (-4)  =  6-4  =  2. 

The  unit  used  in  the  internal  division  is  contained  in 
AB  ten  times.  In  the  case  of  external  division  the  unit 
is  contained  in  AB  two  times.  In  general,  for  internal 
division  the  number  of  divisions  made  in  AB  is  the  sum 
of  the  number  made  in  AM  and  the  number  made  in  MB; 
for  external  division  the  number  of  divisions  made  in 
AB  is  the  difference  between  the  number  made  in  AM ' 
and  the  number  made  in  M'B. 

3.  Divide  sect  CD  harmonically  in   the       ^ D 

ratio  of  5  to  3. 

Sue.     To  find  M  divide  CD  into  8  equal  parts  and  to  find 
M'  divide  CD  into  2  equal  parts. 

4.  Divide  a  given  sect  harmonically  in  the  ratio  of  7  to   5. 
into  how  many  parts  must  it  be  divided? 

5.  If  both  the  interior   and  exterior  angles   at   a  vertex   of 
a  triangle  are  bisected,  the  opposite  side 

is  divided  harmonically  by  the  bisectors.    A~" 

6.  Divide   harmonically   sect  AB   in 

th«s  ratio  of  c  to  d.  ~ — 


110  PLANE  GEOMETRY 

PROPOSITION  V. 

272.  THEOREM.  //  several  lines  are  drawn 
parallel  to  the  base  of  a  triangle  intersecting  the 
sides,  the  corresponding  segments  of  the  sides 
form  a  continued  proportion. 


Given  A  OMN  with  lines  parallel  to  the  base  MN  cut- 
ting the  sides  into  the  sects  a,  b,  c,  d,  e  and  a',  &',  c',  d', 
e' ,  etc  ,  respectively. 

a      b       c      d      c 
To  Prove  -=-=-=-•  =  ->  etc. 
a'      &       c       a      e 

Proof.     SUG.     1.     -  =  -•     Why? 
a       b 

2.  a-  =  ^±.     Why? 
a'     a'+b' 

3.  -^ =~.     Why? 

a'+b'      c' 

4.  Complete  the  demonstration. 
Therefore— 

273.  SIMILAR  POLYGONS.     Polygons   which    are    mu- 
tually equiangular    (§  156)    and  which  have  their  cor- 
responding sides  proportional  are  similar  polygons. 

274.  HOMOLOGOUS.     In  similar  polygons  those  points, 
lines,  and  angles  which  are  similarly  situated  are  homol- 
ogous.     In  similar  triangles  the  homologous  sides  are 


PROPORTION  111 

those  lying  opposite  equal  angles  and  the  equal  angles 
are  those  lying  opposite  homologous  sides. 

275.  '  RATIO  OF  SIMILITUDE.  In  similar  polygons  the 
ratio  of  similitude  is  the  ratio  of  any  two  homologous 
sides.  r> 


D' 


A'1 


The  polygons  P  and  P'  are  similar,  provided 
L  A  =  L  A',      L  B  =  L  B',      L  C  =  L  C',     etc.,     and 

AB      BC       CD  AB 

—  '  etc.     Any  one  of  the  equal  ratios  —  — » 
A'B'    B'C'     C'D'  A'B' 

etc.,  may  be  taken  as  the  ratio  of  similitude. 

276.  From  the  definition  of  similar  polygons,  it  fol- 
lows that,  if  two  polygons  are  known  to  be  similar,  the 
homologous  angles  are  equal  and  the  homologous  sides 
are  proportional. 

1.  In  the  similar  &  ABC  and  A' B'C',  if  L  A  =  L  A',  etc., 
which  sides  are  homologous?  If  the  sides  AB  and  A'B',  etc.  are 
homologous,  which  angles  are  equal? 

277.  Polygons  may  be  mutually  equiangular  but  may 
not  have  their  sides  proportional  or  they  may  have  their 
sides  proportional  without  being  mutually  equiangular. 
The  first  condition  is  illustrated  by  the  rectangle  and  the 
square.     The  second  condition  is  illustrated  by  a  rhom- 
bus and  a  square  or  by  a  rectangle  and  a  rhomboid  if 
the  sides  are  proportional. 

It  will  be  established  later  that  triangles  form  an  ex- 
ception to  the  above  statement,  in  that  if  either  condition 
of  the  definition  of  similarity  applies  the  other  is  a  neces- 
sary consequence. 


112  PLANE  GEOMETRY 

PROPOSITION  VI. 

278.     THEOREM.     Two  triangles  which  are  mu- 
tually equiangular  are  similar. 


ABC     and     A'B'C'     with      Z  A  =  Z  A', 
Z  £  =  Z  B',     Z  (7  =  Z  C'. 
To  Prove  A  ABC  and  A'B'C1'  similar. 
Proof.     SUG.     1.     What  part  of  the  definition  of  sim- 
ilar triangles  remains  to  be  proved? 

2.  Place    &  A'B'C'    upon    A  ABC   so 
that  A'  falls  on  A,  B'  on  AB  at  M  and  6"  on  AC 
at  tf.    Can  this  be  done  ?    Why  do  it  ? 

3.  MN\\BC.    Why? 


4.  .    Why?     ,.. 
Al/     AN  A'B'    A'C' 

5.  What  is  yet  to  be  proved  ? 

6.  Place  A  A'B'C'  upon  A  ABC  with 
B'  on  B,  etc.    Why? 

7.     What    ratios    can    here    be    proved 

equal  ?    Give  all  the  steps. 

1  P 

8.     Compare    the    three    ratios  —  —  » 

yl  x> 

AC.  _AC 
B'C''  A'C' 
Therefore  — 


PROPORTION  113 

279.  COR.  I.     Two  triangles  are  similar  if  two  angles 
of  one  are  equal  respectively  to  two  angles  of  the  other. 

280.  COR.  II.     Two  right  triangles  are  similar  if  an 
acute  angle  of  one  equals  an  acute  angle  of  the  other. 

281.  COR.  III.     Two    triangles   arc   similar   if   they 
are  each  similar  to  the  same  triangle. 

Given  sect  wi.       To  %,* 

divide  m  into  segments  w  ~~..-' 


proportional  to  a,  b,  c, 

0^1 „ 

Sue.    At  one  extrem- 
ity O  of  m  draw  any  line  oblique  to  m  and  on  this  line  from  0  lay 
off  in  order    the  given  sects  a,  b,  c,  etc*    Why?      Complete   the 
construction    by    reproducing  the   conditions     of     proposition    V 
Verify  the  results. 

PKOPOSITION  VII. 

282.     PROBLEM.     To    construct    upon    a    given 
line  a  triangle  similar  to  a  given  triangle. 


E. G 

Given  A  ABC  and  the  line  segment  EG. 
To  Construct  upon  EG  a  triangle  similar  to  A  ABC. 
SUG.     Use  Prop.  XXXV  and  make  the  required 
construction. 

1.  Draw  a  tangent  to  a  given  circle  that  shall  make  a  given 
angle  with  a  given  line. 

2.  Two  isosceles  triangles  are  similar  if  the  vertex  angles  of 
the  two  are  equal. 

3.i       Divide  a  sect  into  segments  proportional  to  three  or  more 
given  sects. 

4.       To  measure  the  height  of  a  nearby  object,  as  EF,  lie  upon 


114 


PLANE  GEOMETRY 


the  ground  in  such  a  position,  AC,  that  the  uppe~  end,  B,  of  a  pole 
of    known    length    placed    vertically    be-  F 

tween  the  feet  will  appear  in  a  line  with 
the  point  F.  The  distances  AC,  CB 
and  AE  are  known  or  easily  measured. 
How  may  EF  be  determined? 

A- 


As  an  illustration  of  the  preceding  problem,  a  woodsman  in 
determining  the  height  of  trees  uses  a  pole  as  long  as  his  own 
height.  What  one  distance  will  he  need  to  measure?  Why? 

5.  How  may  shadows  be   used   to   determine   the   height    of 
objects? 

6.  Determine   as  accurately   as   possible   the  height   of   some 
point  on  the  school  building. 

NOTE.  Not  more  than  two  should  work  together,  the  results 
being  compared  in  class.  D 

7.  To  measure  a  given  height  ED. 

Sue.  Set  up  a  pole  parallel  to  ED 
at  some  convenient  point  as  C  and 
while  one  person  sights  from  a  point 
A  to  D  let  a  second  person  move  a 
card  upon  the  pole  until  a  point  B  is 
found  on  the  pole  in  the  line  AD. 
Make  the  required  measurements  and 
determine  the  height  ED. 

8.  It  is  desired  to   find  the  distance  AB  indirectly.     If  DE 
is  parallel  to  AB  what  measurements  should  be  made? 

9.  A   triangle   has   two   angles   of   69°    and  57°    respectively, 
the  included  side  being  26  rods  in  length.     The  length  of  the  two 
other  sides  is  required. 

Sue.  Construct  on  paper  a  triangle  similar  to  the  given 
triangle,  using  the  protractor  for  the  construction  of  the 
angles.  Measure  the  three  sides.  From  this  data  deter- 
mine the  desired  distances  for  the  given  triangle. 

State  the  various  methods  thus  far  used  for  the  indirect 
measurement  of  distances.     Which  is  the  easiest? 

10.  To  construct  sects  .01,  .02,  .03,  etc.,  to  .09  inches  in  length 
Divide  a  segment  one  inch  long  into   10  equal  parts  and  at  one 


PROPORTION 


115 


extremity  erect  a  perpendicular  .1"  in  length.  Connect  the  other 
extremities  of  these  two  sects  forming  a  right  triangle.  The  per- 
pendiculars to  the  original  sect  at  the  points  of  division  terminated 
by  the  hypotenuse  are  of  the  required  lengths.  Give  the  reasons 
for  each  step.  Take  upon  the  dividers  .03,  .05,  .07  of  an  inch. 

11.      To    construct    a    diagonal    scale  ___T__r_T-~rT~TT~|.r 

by  which  any  segment  may  be  meas-  "rT^r~rT^l  '  '  ' 
ured  in  tenths  and  hundredths  of  an  inch.  The  construction 
should  be  made  on  cardboard  and  preserved  for  future  use, 
if  the  pupil  has  not  already  purchased  a  diagonal  scale.  Con- 
struct a  square  on  a  one  inch  segment,  dividing  each  side  into 
tenths.  Connect  one  vertex  of  the  square  with  the  first  point  of 
division  on  the  opposite  side,  and  through  the  remaining  division 
points  of  this  side  draw  lines  parallel  to  the  first  line.  Through 
the  division  points  of  the  second  pair  of  opposite  sides  draw 
parallels  to  the  first  pair.  Read  from  the  scale  .36,  .42,  .73,  .85, 
.92.  Give  authority  for  all  statements  made. 


12.  With  the  dividers  take  .27  inches  from  the  scale  and  on 
some  given  line  lay  off  a  sect  .27  inches  long: 

13.  Construct    a    sect    3.75    inches    long;    3.56    inches;    2.05 
inches. 

14.  With  the  dividers  and  diagonal  scale  measure  the  sects 
a,  &,  c,  d.  a 

15.  Open  a  jointed  two  foot  rule  so  that          * 

the  ends  are  one  foot  apart.    How  long  is  the  d 

sect  which  connects  points  one  inch  from  the  joint?     2  inches?     5 
inches?     9  inches?     Use  §  276. 

1G.  If  the  ends  of  the  rule  are  six  inches  apart,  how  far  apart 
are  the  pair  of  points  marking  divisions  equally  distant  from  the 
joint? 

17.  Open  the  rule  various  distances  as  above  and  determine 
the  distances  between  corresponding  divisions. 

IS.  Work  out  the  same  exercises,  using  a  one  foot  jointed 
rule.  Also  with  a  six  iiK-h  jointed  rule. 


116  PLANE  GEOMETRY 

PEOPOSITION    VIII. 

283.  THEOREM.  //  tiro  triangles  have  an 
angle  of  one  equal  to  an  angle  of  the  other  and 
the  sides  including  the  equal  angles  proportional, 
the  triangles  are  similar. 

A 


B  C     B'  C' 

Given    A  ABC   and    A  A'B'C'    with    ZA  =  A'    and 


AB'     AC' 

To  Prove  A  ABC  ^  A  A'B'C'. 

Proof.  SUG.  1.  What  must  be  proved  in  addition 
to  the  hypothesis  to  make  the  triangles  similar 
according  to  §  278  ? 

2.  Place    A  A'B'C'   upon    A  ABC   so 
that  A'  falls  on  A,  A'B'  on  AB,  and  A'C'  upon 
AC.     Is  this  possible?    Why  do  so? 

3.  Where  do  B'  and  C'  fall? 

4.  B'C'^BC.     Why? 

5.  Compare    Z  B'   with    Z  B.     Com- 
plete the  demonstration. 

Therefore — 

1.  If  triangles  have  their   sides  respectively  parallel  or  per- 
pendicular to  each  other  they  are  similar. 

2.  Let  ABC  and  A'B'C'  be  two  similar  triangles.     AB-1 
ft.,  A'B'  =  14  ft.,  AC  =  5.     Find  the  length  of  A'C'.     If  AB  is 
is' ft.,  AC  and  A'B'  are  11  ft.  and  10  ft.  respectively;    find  the 
length  of  A'C'. 


PROPORTION  117 

PROPOSITION  IX. 

284.     THEOREM.     Two  triangles  are  similar  if 
the  corresponding  sides  are  proportional. 


Given  A  ABC  and  A'B'C'  with  -—  =-= 

A'B'      B'C'      A'C'' 

To  Prove   A  ABC  —  A 'A'B'C'. 

Proof.    SUG.     1.     If  any  angle  of  A  ABC  equals  the 

homologous  angle  of  A  A'B'C'  the  triangles  are 

similar.     Why  ? 

2.     Upon   AB    lay   off   AM   equal   to 

A'B'  and  upon  AC  lay  off  AN  equal  to  A'C'. 

Connect  M  and  N.     A  AMN ^ '  A  ABC.     Why? 

3.  Compare  the  ratios  -    -   and  - 

AB  BC 

,      A'B'       ,  B'C' 

also and • 

AB  BC 

4.  Compare  MN  and  #'<?'.      Auth.? 
A  AMN  =  A  A'B'C'.     Auth. 

5.  A  ABC  — '  A  A'B'C'.     Why? 
Therefore— 

1.  The   sides   of  a  triangle  ^4BC  are  respectively  4,   8,   and 
11  feet.     In  a  similar  triangle,  A'B'C',  the  side  homologous  to  the 
4  foot  side  of  ABC  is  6  ft.     Find  the  two  other  sides  of  A'B'C'. 

2.  In  Prop,  jx  why  n°t  place  A'B'C'  upon  ^4J5(7  as  in  Prop. 
VIII. 


118 


PLANE  GEOMETRY 


PKOPOSITION  X. 

285.  THEOREM.  The  ratio  of  homologous  al- 
titudes of  similar  triangles  is  equal  to  the  ratio 
of  similitude  of  the  triangles. 


Given     A  ABC  --  A  A'B'W    with   — as   the  ratio 

A'B' 

of  similitude,    and   altitude  MA   homologous   to   M'A'. 

„    _  MA       AB 

To  Prove — 

M'A'     A'B' 

Proof.     SUG.     1.     Compare     A  AMB    and    A'M'B'. 

§  280. 

2.  Show  that  MA  and  M'A'  are  ho- 
mologous sides  of  A  AMB  and  A'M'B'. 

3.  Complete  the   demonstration. 
Therefore— 

285.  (1)  COR.  In  similar  triangles  homologous  altitudes 
arc  proportional  to  Hie  bases. 

MA       BC 


SUG. 


Prove  -    —  = 

M'A1     B'C' 


NOTE.  In  deriving  proportions  from  similar  polygons  in  the 
early  study  of  the  subject  it  is  usually  best  to  select  homologous 
sides  for  the  terms  of  each  ratio,  taking  the  antecedent  from  one 
polygon  and  the  consequent  from  the  other.  If  any  other  form 
of  proportion  is  desired  it  can  be  deduced  by  the  theorems  on 
proportion. 


PROPORTION  119 

PROPOSITION  XL 

THEOKEM.  If  two  polygons  are  composed  of  the 
same  number  of  triangles,  similar  ea.cli  to  each  and 
similarly  placed,  the  polygons  are  similar. 


C' 

To  Prove 

Given    polygons  P   and  P' ;  P  composed  of  A    1,  2, 
3    .  .  .  andP'of  1'2'3'...;   Al  —Al',  A2— A2'.    ... 
Proof.    SUG.  1.     What  is  the  test  of  similarity  of  poly- 
gons? 

2,  Compare  <B  and  <B'  •  C  and  C',  etc. 

AB       BC       BC       CD 

3,  Compare  -—,—,-—,-      — , 

A'Bf     B'C'    B'C'    C"D' 
HC  CD 

etc.     To  compare  -       -  and  -      -  relate  each  to 
B'C'  C'D' 

AC 

the  ratio   • 

A'C' 

4,  Apply  the  difinition  §273. 

286  (1).     COR.    Two  similar  polygons  can  be  divided 
into  the  same  number  of  triangles  similar  each  to  each 
and  similarly  placed. 
Use  figure  §286. 

Prove  that  the  polygons  can  be  cut  in  the  same 
number  of  A  ;  that  Ax  - — 'Aj'.     A2  • — 'A2',  etc. 
Proof.     SUG.  1,     A  i— A  /  §  283. 

2,     To  compare  A2    and    A2.    Compare 

ACD  and  A' C'D',  Compare  ratios  -   —-and  ~ 


120  PLANE  GEOMETRY 

To  compare  these  ratios  note  their  common  rela 

BC 

tion  to  the  ratio  - 

B'C' 

1.        Straight  lines  drawn  through  any  point  intercept  propor- 
tional segments  upon  two  or  more  parallel  lines. 


SUG.    Compare  the  ratios 
AB     BC  OB 

'  etc" 


Complete  the  demonstration. 

2.  If  two  or  more  parallel  lines  are 
cut  proportionally  by  a  set  of  secant 
lines,  prove  that  the  secant  lines  pass 
through  a  common  point.  That  is,  given 

etc.,     prove     that     the 


lines  A'A,  B'B,   C'C,   etc.,   pass   through 
a  fixed  point  0. 

SUG.     Let  D  be  the  intersection  point  of  two  of  the  lines  as 
A'A  and  B'B.    Connect  0  with  C  and  extend  DC  to  meet  the 

BC  BC 

second  parallel  at  M.    Compare  the  ratios!    75777  and    ,,,,,  • 

JtJ   C/  ij   JxL 

3.  Given    an   isosceles    triangle    with    vertex   angle    of    120°. 
Prove   that  the  altitude  from  the  vertex   angle  equals   ^   a  leg 
of  the  triangle. 

4.  If  tangents  to  a   circle  be  drawn  at   the  extremities  of  a 
chord,  these  tangents   make   with  each  other  an  angle  which  is 
twice  the  angle  between  the  chord  and  that  diameter  of  the  circle 
drawn  through  an  extremity  of  the  chord. 

5.  Given  sect  A  the  diagonal,  and  sect  B  the  side  of  a  square 
construct  the  square. 

Work  out  through  methods. 

6.  Points  A  and  B  are  on  the  Siame  side  of  line  C.     Find  point 
.V  such  that  AX  and  BX  make  equal  angles  with  C. 


TRIGONOMETRIC   RELATIONS  121 

TRIGONOMETRIC  RELATIONS. 

287.  It  has  been  established  that  in  a  right  triangle 
with   an  angle   of  30°  the   side 

opposite  this  angle  is  one  half 
the  hypotenuse.  This  is  true 
for  every  right  triangle  having 
an  angle  of  30°,  for  all  such 
triangles  are  similar. 

With  a  protractor  carefully  construct  a  right  triangle  with  an 
angle  of  40°,  measure  the  hypotenuse  and  side  opposite,  and  de- 
termine the  ratio  of  the  latter  to  the  former.  Kepeat  the  con- 
struction several  times  and  average  the  results.  Do  the  same 
for  a  right  triangle  having  an  angle  of  50°  j  of  60° ;  and  tab- 
ulate the  results. 

288.  SINE  OF  AN  ACUTE  ANGLE.     The  ratio  of  the 
side  opposite  an  acute  angle  of  a  right  triangle  to  the 
hypotenuse  is  the  sine  of  that  angle.     Taking  the  no- 
tations  from   the   fig.    of    §  287    this    is   written   sym- 
bolically as  sin  A  =  —  and  sin  B  —  — ,  or   c   sin  A  =  a 

c  c 

and  c  sin  B  =  b. 

If  sin  A  =  .5  then  a  =  .5  c.  If  c  =  40  in.,  then  a  =  $ 
of  40  in.  =  20  in. 

NOTE.  The  pupil  should  clearly  understand  that  the  trigono- 
metric ideas  here  explained  are  introduced  as  an  aid  to  the  study 
of  certain  simple  figures  and  that  the  complete  study  of  trigo- 
nometry will  require  more  general  definitions  than  those  here  given. 

1.  In  a  right  triangle  the  sine  of  an  angle  is  .75  and  the 
hypotenuse  is  20  in.     Find  the  length  of  the  side  opposite  this 
angle. 

2.  If  the  hypotenuse  is  20  in.  and  one  side  is  lo  in.  what  is 
the  sine  of  the  angle  opposite  this  side? 

NOTE.  Tables  have  been  made  in  which  the  sines  of  angles  for 
every  degree  and  minute  from  0°  to  90°  are  recorded.  A  portion 
of  such  a  table  for  intervals  of  one  degree  is  to  be  found  on  p.  125 


122  PLANE  GEOMETRY 

1.  How  many  degrees  in  the  angle  found  in  Ex.  2  p,  121. 

2.  The  hypotenuse   of   a    triangle   with   an   angle    of   40°    is 
60  in.     How  long  is  the  opposite  side? 

3.  If  the  side  opposite  an  angle  of  35°  is  34  in.,  how  long 
is  the  hypotenuse? 

4.  The  hypotenuse  is  75  in.,  how  many  degrees  in  each  angle 
if  one  side  is  43  in? 

7.  A  flag  staff  is  100  feet  high.  How 
long  a  rope  is  needed  to  reach  from  the  top 
to  a  point  on  the  ground  at  which  the  angle 
subtended  by  the  pole  is  40°? 

289.  COSINE  OF  AN  ACUTE  ANGLE.     The  ratio  of  the 
side  of  a  right  triangle  adjacent  to  an  acute  angle  to 
the  hypotenuse  is  the  cosine  of  that  angle. 

6.  If  a  right  triangle  has  an  angle  of  60°,  it  has  already  been 
shown  that  the  ratio  of  the  adjacent  side  to  the  hypotenuse  is  1/{». 
Draw  such  a  triangle  and  verify  the  statement  by  measurement. 

7.  Draw  a  right   triangle  with   an  angle  of  40°.     Compute 
the  cosine  by  measurement  and  check  the  result  by  the  table.     Do 
the  same  for  50°  and  60°. 

8.  In   the    accompanying    figure   the 
relations    involving    cosines    are    written 

&  a 

cos  A  =  ~~  and  cos   B  —  ~~>  or  c'cos  A=b 
c  c 

and  c  •  cos  B  =  a.  If  cos  A  =  .5  then  b  =r.5  c. 

If  c  =  40   in.    then  b  =    i  of  40   in.  =  20  b 'c 

A 

in. 

9.  One  angle  of  a  right  triangle  is    45°    and  the  hypotenuse 
is  60  ft.     How  long  is  the  side  adjacent  to  the  given  angle?    How 
long  is  the  side  opposite? 

10.  If  the  adjacent  side  is   20   in.  and  the  hypotenuse  is  63 
in.  what  is  the  cosine?     How  many  degrees  in  the  angle? 

11.  If  the  angle  is  33°   and  the  adjacent  side  is  37   ft.  how 
long  is  the  hypotenuse?     How  long  is  the  side  opposite? 

290.  THE    TANGENT    OF    AN    ACUTE    ANGLE.     In    a 
right  triangle  the  ratio  of  the  side  opposite  an  acute 
angle  to  the  side  adjacent  is  the  tangent  of  that  angle. 


TRIGONOMETRIC   RELATIONS  123 

In  the  figure  of  §  289  the  relations  involving  tangents 

are  written  tan  A  =  -  and  tan  B  =—   or  Irtan  A  =  a 

I)  a 

and  a -tan  B=b. 

1.  Construct  a  right  triangle  with  an  angle  of  30°.     Meas- 
ure the  two  sides  and  compute  the  tangent  of  30°. 

2.  What  is  the  height  of  a  flag  staff  if  the  angle  subtended 
by  the  staff  at  a  distance  of  75  ft.  is  35°? 

3.  In  a  right  triangle  the  sides  are  3  in.  and  4  in.  respect- 
ively.    What  are  the  tangents  of  the  acute  angles? 

4.  A  tower  75  ft.  high  stands  beside  a  river.     The  line  from 
the  top  to  a  point  on  the  opposite  bank  makes  an  angle  of  5i)0 
with  the  tower.     How  wide  is  the  river? 

291.  ANGLE  OF  ELEVATION.     If  at  any  point   in  a 
horizontal  or  level  line  a  line  be   drawn  to  a   second 
point   above  the  horizontal,  the       i 

angle  thus  formed  is  the  angle 
of  elevation  of  the  second  point. 
If  BC  is  a  horizontal  or  level 
line  the  angle    B    is  the  angle  of  elevation  of  point  A 
from  B. 

292.  ANGLE  OF  DEPRESSION.     If  at  any  point  in  a 
horizontal  line  a  line  be  drawn  to  a  second  point  below 
the  horizontal  the  angle  thus  formed  is  the  angle  of  de- 
pression of  the  second  point. 

L  DAB  is  the  angle  of  depression  of  B  from  A. 

293.  HORIZONTAL  ANGLES.    Angles  in  a  horizontal  or 
level  plane  are  horizontal  angles. 

294.  VERTICAL  ANGLES.     Angles  of  elevation  and  de- 
pression are  vertical  angles  as  distinguished  from  hor- 
izontal angles. 

5.  From   the  top  of  a  tower  the  angle   of  depression   of  a 
vessel  at  sea  is  15°.    If  the  tower  is  87  ft.  high  hpw  far  away  is 
the  vessel? 


124 


PLANE  GEOMETRY 


1,  Points  B  and  C  are  on  opposite 
banks  of  a  river.  Line  BC  along  the  bank  is 
150  ft.  long,  Z  A  is  43  ,  L  C  is  90°.  How  Avide 
is  the  river  ? 


295.  In  any  triangle  the  sides  of  two  acute  angles 
have  the  same  ratio  as  •  he  sides  opposite. 

Given  A  ABC  with  sides  a,  b,  c 
and  acute  angles  B  and  C.  A 

m     _  sin  B      b 

To  Prove          -— -. 

sin  C      c 

Sue.     Express  sin  B  and  sin  C. 
Complete  the  demonstration. 

NOTE:  This  is  a  very  important  proposition  as  it  gives  certain 
relations  connecting  the  sides  and  angles  of  triangles  other  than 
right  triangles.  It  is  proved  here  only  for  the  case  in  which  the 
two  angles  concerned  are  acute.  It  will  be  proved  in  trigonometry 
without  exception  and  is  known  as  the  law  of  sines. 

[2.  In  &ABC,  A  =15°,  C=76°,  and  side  c  -  60  in.  Find 
the  other  sides  of  the  triangle  and  L  B. 

sin  A         a ,   .  .    sin  A 


SUG,    1. 


—  —  and  .'.  a  —  c 


sin  C          c  sin  C 

2.  Since  all  angles  of  this  triangle  are  acute  this 

law  may  be  likewise  used  for  angles  A  and  B. 

3.  Consult  the  table  and  complete  the  problem. 

3.  From   a   certain   point   the   eleva- 
tion of  the  top  of  a  church  tower  is  43°. 
From  a  point  100  ft.  nearer  the  base  of 
the  tower  the  angle  of  elevation  is  55°. 
Find  height   of  the  tower. 

Sue.    Find  sect  BC  from  A  ABC 
and  then  CD  in  A  BCD.  A- 

4.  Knowing  the   distance   AC  to   be  8  mi.,  the  angle  A  to  be 
40°,  and  the  angle  C  to  be  82°  j  find  the  inaccessible  distance  AB 
across  the  lake. 


TRIGONOMETRIC   RELATIONS 


12s 


296.     TRIG.  TABLE. 

The  Sines,  Cosines  and  Tangents  of  Acute  Angles. 


Degrees 

Sin. 

Cos. 

Tan  . 

Degrees 

Sin. 

Cos. 

Tan. 

1 

.0175 

.9998 

.0175 

46 

.7193 

.6947 

.0355 

2 

.0349 

.9994 

.0349 

47 

.7314 

.6820 

.0724 

3 

.0523 

.9986 

.0524 

48 

.7431 

.6691 

.1106 

4 

.0698 

.9976 

.0699 

49 

.7547 

.6561 

.1504 

5 

.0872 

.9962 

.0875 

50 

.7660 

.6428 

.1918 

6 

.1045 

.9945 

.1051 

51 

.7771 

.6293 

.2349 

7 

.1219 

.9925 

.1228 

52 

.7880 

.6157 

.2799 

8 

.1392 

.9903 

.1405 

53 

.7986 

.6018 

.3270 

9 

.1564 

.9877 

.1584 

54 

.8090 

.5878 

.3764 

10 

.1736 

.9848 

.1763 

55 

.8192 

.5736 

.4281 

11 

.1908 

.9816 

.1944 

56 

.8290 

.5592 

.4826 

12 

.2079 

.9781 

.2126 

57 

.8387 

.5446 

.5399 

13 

.2250 

.9744 

.2309 

58 

.8480 

.5299 

.6003 

14 

.2419 

.9703 

.2493 

59 

.8572 

.5150 

.6643 

15 

.2588 

.9659 

.2679 

60 

.8660 

.5000 

.7321 

16 

.2756 

.9613 

.2867 

61 

.8746 

.4848 

.8040 

17 

.2924 

.9563 

.3057 

62 

.8829 

.4695 

.8807 

18 

.3090 

.9511 

.3249 

63 

.8910 

.4540 

1.9626 

19 

.3256 

.9455 

.3443 

64 

.8988 

.4384 

2.0503 

20 

.3420 

.9397 

.3640 

65 

.9063 

.4226 

2.1445 

21 

.3584 

.9336 

.3839 

66 

.9135 

.4067 

2.2460 

22 

.3746 

.9272 

.4040 

67 

.9205 

.3907 

2.3559 

23 

.3907 

.9205 

.4245 

68 

.9272 

.3746 

2.4751 

24 

.4067 

.9135 

.4452 

69 

.9336 

.3584 

2.6051 

25 

.4226 

.9063 

.4663 

70 

.9397 

.3420 

2.7475 

26 

.4384 

.8988 

.4877 

71 

.9455 

.3256 

2.9042 

27 

.4540 

.8910 

.5095 

72 

.9511 

.3090 

3.0777 

28 

.4695 

.8829 

.5317 

73 

.9563 

.2924 

3.2709 

29 

.4848 

.8746 

.5543 

74 

.9613 

.2756 

3.4874 

30 

.5000 

.8660 

.5774 

75 

.9659 

.2588 

3.7321 

31 

.5150 

.8572 

.6009 

76 

.9703 

.2419 

4.0108 

32 

.5299 

.8480 

.6249 

77 

.9744 

.2250 

4.3315 

33 

.5446 

.8387 

.6494 

78 

.9781 

.2079 

4.7046 

34 

.5592 

.8290 

.6745 

79 

.9816 

.1908 

5.1446 

35 

.5736 

.8192 

.7002 

80 

.9848 

.1736 

5.6713 

36 

.5878 

.8090 

.7265 

81 

.9877 

.1564 

6.3138 

37 

.6018 

.7986 

.7536 

82 

.9903 

.1392 

7.1154 

38 

.6157 

.7880 

.7813 

83 

.9925 

.1219 

8.1443 

39 

.6293 

.7771 

.8098 

84 

.9945 

.1045 

9.5144 

40 

.6428 

.7660 

.8391 

85 

.9962 

.0872 

11.4301 

41 

.6561 

.7547 

.8693 

86 

.9976 

.0698 

14.3007 

42 

.6691 

.7431 

.9004 

87 

.9986 

.0523 

19.0811 

43 

.6820 

.7314 

.9325 

88 

.9994 

.0349 

28.6363 

44 

.6947 

.7193 

.9657 

89 

.9998 

.0175 

57.2900 

45 

.7071 

.7071 

1  .0000 

126  PLANE  GEOMETRY 

PROPOSITION    XII. 

297.  THEOREM.  In  the  same  circle  or  in  equal 
circles  central  angles  are  proportional  to  the  arcs 
they  intercept. 


Given  OC  =  OC"  with  central  angles  C  and  C"  in- 
tercepting the  arcs  AB  and  A'B'  respectively. 
Z  C       arc  AB 


To  Prove 


Z  C'     arc  A'B' 


Z  C 
Proof.     SUG.     1.      To    obtain    the    ratio  the 

ZG" 

angles  must  be  measured.  Auth.?  Suppose  the 
unit  angle  e  to  be  contained  in  Z  C  exactly  m 
times  and  in  Z  C'  exactly  n  times. 

2.  ...   ^  =  ™      Why? 

Z  C'        n 

3.  To     obtain    the    ratio  the 

A'B' 

arcs  must  be  measured.  For  a  unit  arc  take  the 
arcs  intercepted  by  the  unit  angles.  Auth  ?. 

4.  How  many  of  these  unit  arcs  in 
ABt    Why?    In  A'B'6!    Why? 

5.  What  then  is  the  ratio  of   AB   to 
A'B"! 

/  f1  4  "D 

6.  Compare  the  ratios  -    -  and  - 

L  C'         A'B' 

Therefore — 


PROPORTIONAL    LINES 


127 


NOTE.  This  demonstration  does  not  cover  the  case 
in  which  the  two  angles  are  incommensurable.  See  note 
on  §  251. 

298.  DEGREE  OP  ARC.     The  arc  intercepted  by  a  cen- 
tral angle  of  one  degree  is  a  degree  of  arc. 

299.  COR.     The    number  of  degrees   of  angle   in   a 
central  angle  equals  the  number  of  degrees  of  arc  in  the 
intercepted  arc. 

For:     If  central   angle  A   intercepts  arc  a  then  §297 
LA .__   arg  a 
I1 


But  these  two  ratios  are  respectively 


1°  of  arc 
the  number  of  degrees  in  the  angle  and  the  arc. 

This  important  theorem  is  usually  stated  thus:      A 
central  angle  is  measured  by  its  intercepted  arc.    §  297. 

If  A  and  a  are  notations  for  the  angle  and  arc  the 
relation  between  them  expressed  in  symbols  is 
Z  .4  =  area.     (§50.) 

1.  Are  all  degrees  of  arc  the  same  length! 

2.  How   many   degrees    in   a  circle'?     In  a  semicircle?     Jn  a 
quadrant  ? 

PROPOSITION  XIII. 

300.     THEOREM.    An  inscribed  angle  is   meas- 
ured by  one  half  its  intercepted  arc. 


Given  O  C  with  the  inscribed   Z  ABD  intercepting 
the  arc  AD. 
To  Prove  Z  ABD  =c  \     arc  AD. 


128  PLANE  GEOMETRY 

Proof.    There  are  three  cases. 

CASE  I.     One  side  of  the  angle,  BD,  is  a  diameter. 

SUG.     1.  Connect    A    and    C.      Z  m  —  2  Z  B. 
Why? 

2.  .'•  LB  =  \    Zm. 

3.  •'•  Z  £n:  I   arc^LZ).     Why? 

CASE  II.     The  center  of  the  circle  lies  between  the 
sides  of  the  angle. 

Sue.     1.     Z  m  is  measured  by  what?    Why? 

2.  Z  ?i  is  measured  by  what?     Why? 

3.  L  'R  is  measured  by  what  ?     Why  ? 
CASE   III.     The   center   of   the   circle   is  without  the 

angle. 

Proof  is  left  to  the  pupil. 
Therefore— 

NOTE:  This  theorem  may  also  be  stated  thus:  The  numbei 
of  degrees  in  an  inscribed  angle  is  one-half  the  number  in  the 
intercepted  arc  and  the  same  change  in  wording  may  be  made 
throughout  the  proof. 

301.  COR.  I.     An  angle  inscribed  in  a  semicircle  is  a 
right  angle. 

302.  COR.  II.     A  segment  in  which  a  right  angle  is 
inscribed  is  a  semicircle. 

303.  COR.  III.     All  angles  inscribed  in  the  same  or 
equal  segments  are  equal. 

304.  COR.  IV.     An    angle    inscribed   in    a    segment 
greater  than  a  semicircle  is  acute:  in  a  segment  less  than 
a  semicircle  is  obtuse. 

1.      Construct  a  right  triangle  by  means  of 
Cor.  I. 

2.1     In   O  C  with  diameter  AB  prove 
/  m  =  L  n. 

3.     If  DE  is  a  diameter  of  0  C  which  angle 
is  the  greater,  w  or  n?  n  or  of 


PROPORTIONAL    LINES  129 

PROPOSITION  XIV. 

305.  PROBLEM.  To  construct  a  right  triangle 
when  the  hypotenuse  and  an  acute  angle  are 
given. 


A —  — 'B  A" 

Given  sect  AB   as  the  hypotenuse   and    L  A   as  an 
aciite  angle  of  a  right  triangle. 
To  Construct  the  triangle. 

SUG.     Construct  the  given  angle  at  point  A  of 
AB.     At  B  erect  a  1.     Complete  the  problem. 

1.        To  construct  a  perpendicular  to  a  given 
line  c  from  a  point  M  on  that  line.     See  §  301. 


Sue.  With  a  point  C  not  on  the 
line  as  center  and  with  CM  as  radius 
draw  a  circle  cutting  line  c  in  a  second  point  A.  Draw 
AC  meeting  the  circle  again  in  B.  Then  MB  is  the  required 
line.  Why? 

2.  From  a  given  point  B  not  on  a  line  c  draw  a  perpendicular 
to  c.     See  §  301. 

Sue.  Draw  any  oblique  line  through  B  meeting  line  c 
in  point  A.  On  AB  as  diameter,  construct  a  circle  meeting 
c  in  point  M.  The  required  line  is  MB.  Why? 

3.  Draw    two    lines    from    A    and    B  N 
meeting  on  the  line  MN  so  as  to  form   a 

right  angle.     Is  more  than  one  construction 
possible? 

4.  Measure    the    height    of   your   school      » 

building   using    two  angles  of  elevation  and 

n  distance.     See  Ex.  4,  P.  124. 

5.  With  the   four  vertices   of   a  square   as  centers   and  with 
radii  equal   to   one-half  a  side   of  the   square   draw   four   circles. 
Show  that   one  circle  can  be   drawn  which  is  externally  tangent 
to  the  four  circles. 


130  PLANE  GEOMETRY 

PROPOSITION  XV. 

306.  THEOKEM.  An  angle  formed  by  two  in- 
tersecting chords  is  measured  by  one  half  the 
sum  of  the  intercepted  arcs. 


Given  O  O  with  chords  CE  and  BD  intersecting  at  A. 
To  Prove  Ln^\   (arc  BC  +  arc  DE). 
Proof.     SUG.     1.     Draw   CD.      L  n  =  L  C  +  Z  D. 
Why? 

2.  "What  is  the  measure  of   Z  (7?  of 
of  ZD? 

3.  Complete  the  proof. 
Therefore— 

1.  Show  two  methods  of  finding  the  center  of  an  equilateral 
triangle. 

NOTE:     The  center  of  a  polygon  is  that  point  which  is  equidis- 
tant from  the  vertices.     Some  polygons  do  not  have  centers. 

2.  On  top   of    a  hill  '200  ft.  high  is  a 
tower.        From  one  point  in  the  level  plane 
at  the  foot  of  the  hill  the  elevation  of  the 
top  of  the  tower  is  29°.        At  a  point  200 
ft.  nearer  the  foot  of  the  kill  the  elevation 
is  40°.     How  high  is  the  tower? 

3.  Inscribe  a  triangle  in    a    given    circle    similar    to    a    given 
triangle. 

4.  Circumscribe  a  triangle  about  a  given    circle    similar    to    a 
given  triangle. 

5.  If  two  circles  are  tangent  internally  and  through  th«  point 
of  tangency  a  line  is  drawn,  the  chords  intercepted  by  the  circu-s 
are  proportional  to  the  radii  of  the  circles. 


PROPORTIONAL   LINES  181 

PROPOSITION  XVI. 

307.  THEOREM.  An  angle  formed  by  a  tan- 
gent and  a  chord  drawn  from  the  point  of  contact 
is  measured  by  one  half  the  intercepted  arc. 


Given  the  O  with   L  m  formed  by  the  tangent  AB 
and  the  chord  AC. 
To  Prove   Z  ra  m  J  arc  AEC. 

Proof.     SUG.     1.     Through  C  draw  a  line  parallel  to 
AB,  as  CD. 

2.  Compare    A  m   and   n ;   arcs   AEC 
and  AFD.     Auth. 

3.  Complete  the  proof. 
Therefore — 

1.  Two   circles   are   tangent  internally.     Two   lines  are  drawn 
from  the  point  of  tangency  through  the  extremities  of  a  diameter 
of  one  circle.     Prove  that   they  intersect  the  other  circle  in  the 
extremities  of  a  diameter. 

2.  Prove  Ex.  1  above  if  the  circles  are  tangent  externally. 

3.  AB  and  CD  are  two  chords  of  a  circle  intersecting  in  the 
point  0.     Prove  A  AOD  and  A  COB  mutually  equiangular.     Prove 
the  same  for  A  AOC  and  A  BOD. 

4.  What  is  the  locus  of  the  vertex  of  the  right  angle  of  a 
right  triangle  with  a  given  sect  as  hypotenuse?     (§302.) 

5.  Construct  a  triangle  having  a  given 
base,  a  given  altitude,  and  a  right  vertex 
angle. 

6.  Prove  Prop.   XV  from  the  accom- 
panying figure  in  which  EM  \  BD. 


132 


PLANE  GEOMETRY 


PROPOSITION  XVII. 

308.  THEOREM.  An  angle  formed  by  two 
secants,  a  secant  and  a  tangent,  or  by  two  tan- 
gents is  measured  by  one  half  the  difference  of 
the  intercepted  arcs. 


Given  O  0  with  I.  two  secants  AD  and  AE ;  II.  secant 
AD  and  tangent  AB  •  III.  two  tangents  AB  and  AC. 
To  Prove  I.     Z  A  m  •*•  (Arc  DE  —  etrcBC) ; 

II.  LA^\  (Arc  BD  —  arc  EC)  • 

III.  Z  A  m  |  ( arc  SMC  —  arc  BNC ) . 

Proof.     CASE  I.     SUG.  1.     Compare    Z  A   with    Z  m 
and  Z  #. 

2.         What  is  the  measure  of 
Z  m,  Z  #,  Z  A  ? 

CASE  II.     Left  to  the  pupil. 
CASE  III.    Left  to  the  pupil. 
Therefore — 

1.  All  angles  inscribed  in  the  same  segment  are  equal. 

2.  In  the  same  or  in  equal  circles  an     angle  inscribed    in    the 
smaller  of  two  segments  is  greater  than  an  angle  inscribed  in  the 
larger  segment. 

3.  What  is  the  locus  of  the  vertex  of  a     triangle    having    a 
given  base  and  altitude? 

4.  A  chord  is  met  at  one  extremity  by  a  tangent  forming  an 
angle  of  75°.     How  many   degrees  in  the  arc  that  is  subtended 
by  the  chord? 


PROPORTIONAL    LINES  133 

PROPOSITION  XVIII. 

309.  THEOREM.  An  angle  formed  by  two  lines 
either  or  both  of  which  may  be  a  secant  or  a  tan- 
gent to  a  circle  is  measured  by  one  half  the  sum 
of  the  intercepted  arcs. 


Proof.     SUG.  1.     Is  the  theorem  true  if  the  lines  in- 
tersect at  the  center  ?    Why  ? 

2.  Is  the  theorem  true  if  the  intersec- 
tion be  anywhere  within  the  circle?     Why? 

3.  If     the     intersection     points     ap- 
proach nearer  and  nearer  to  the  circle  what 
happens  to  one  of  the  intercepted  arcs?     Sup- 
pose that  when  the  point  of  intersection  is  on 
the  circle  this  arc  be  considered  as  zero.     Is  the 
theorem  true  for  this  case? 

4.  A  comparison  of  the  figures  shows 
that  as  0  moves  from  a  position  within  the  O 
to  a  position  on  the  circle  the  points  0,  A  and  C 
come   together   and  as   0   passes   without  the 
circle  they  again  separate  but  with  this  dif- 
ference, that  A  and  C,  points  in  which  the  lines 
BA  and  DC  meet  the  circle,  in  the  third  figure 
have  their  relative  positions  reversed,  so  that 
the  arcs  AC  in  the  first  and  third  figures  are 
opposite  in  direction  and  if  the  idea  of  positive 
and  negative  lines  be  introduced  the  arc  —  CA 
which   enters  the   formula   for  the  third  case 
may  be  written  as  +  AC.  §269. 


134  PLANE  GEOMETRY 

5.     For  each  case  then  the  measure  of 
the  angle  at  0  is  J  (arc  DB  +  arc  AC). 
The  pupil  should  complete  the  demonstration  for  the 
case  involving  one  or  two  tangent  lines. 

310.  CONTINUITY.     A    theorem    can    sometimes    be 
stated  in  such  a  general  way  as  to  cover  two  or  more 
particular  theorems.     Especially  is  this  true  in  geometry 
when  a  distinction  is  made  as  to  the  direction  of  lines 
represented  algebraically  by  the  use  of  positive  and  nega- 
tive quantities.     Such  a  theorem  is  that  of  §  309  which 
is  proved  by  the  principle  of  continuity. 

1.  In  making  a  pattern  for  a  certain  casting  it  is  necessary 
to   construct  a  true  semicircle.     How  may  this  be  done  witk  a 
carpenter 's  square  ? 

2.  At  a  given  point  on  a  circle  construct  a  tangent  to  the 
circle. 

Sue.  If  a  line  were  drawn  from  the 
center  C  to  the  point  A  what  relation 
would  it  bear  to  the  tangent  through  A? 
Complete  the  construction. 

PROPOSITION  XIX. 

311.  PROBLEM.    From  a  given  point  without  a 
circle  to  construct  a  tangent  to  the  circle. 


Given  O  C  with  point  A  without. 
To  Construct  a  tangent  to  O  C  from  A. 

SUG.     1.     Connect  C  and  A.     The  problem  is 
to  determine  the  point  of  tangency. 


PROPORTIONAL    LINES 


135 


2.  What    relation    exists    between    the 
tangent  and  the  radius  at  the  point  of  tangency  ? 

3.  What  is  the  locus  of  a  point  with 
respect  to  AC  satisfying  this  condition? 

4.  Complete   the  construction.     §   302. 

5.  How  many  such  tangents  are  there? 
Why? 

6.  Make  a  second  construction  drawing 
only  such  portions  of  the  auxiliary  lines  as  are 
necessary. 


312.     FIND  THE  Locus : 

(1)  Of  centers  of  circles  having  a  given  radius 
and  tangent  to  a  given  line. 

(2)  Of  centers  of  circles  tangent  to  a  given 
line  at  a  given  point. 

(3)  Of  centers  of  circles  having  a  given  radius 
and  tangent  to  a  given  circle. 

(4)  Of  centers  of  circles  tangent  to  two  inter- 
secting lines. 

(5)  Of  centers  of  circles  having  a  given  radius 
and  passing  through  a  given  point. 

(6)  Of  centers  of  circles  passing  through  two 
fixed  points. 

1.  Prove  Prop.  XVII  by   drawing  from  the  point  D  in  the 
accompanying  figures  a  line  DM  1 1  AC. 

2.  Two  circles  intersect  in  the  points  A   and  D.     Lines  AB 
and  AC  are  diameters.     Prove  that  B,  D,  and  C  lie  in  a  straight 
line. 


136 


PLANE  GEOMETRY 


1.  The  sum  of  the  distances  from  points  in  the  base  of  an 
isosceles  triangle  to  the  legs  is  constant. 

2.  Given  three  non-parallel  lines  unlimited  in  length.     Find 
points  in  one  of  them  equally  distant  from  the  two  others.     How 
many  such  points  are  there? 

3.  Two  circles  'intersect  in  the  points  A  and  D.     Sects  AB 
and  AC  are  chords  of  the  two  circles.     Points  B,  I),   C  are  in  a 
straight  line.     Prove  that  the  chords  are  diameters. 

4.  One  side  of  an  equilateral  inscribed  hexagon  is  equal  to 
the  radius  of  the  circle. 

5.  Inscribe    an    equilateral    triangle    in    a 
circle  and  prove  that  the  radius  perpendicular 
to  a  side  is  bisected  by  the  side. 

Sue.     OABD  is  a  /~7.     Why? 

6.  Two   chords   drawn  from   a  point  on   a 
circle    are    inversely    proportional    to    the    seg- 
ments of  the  chords  included  between  the  tan- 
gent at  their  common  point  and  a  line  parallel 
to  this  tangent. 


To  PROVE     *»  =  4*.. 
AE      AC 


See  note,  §285  (1). 


7.  The  locus  of  the  middle  points  of  all  chords 
which  pass  through  a  given  point  is  a  circle  the 
diameter  of  which  is  the  line  joining  the  given  point 
and  the  center  of  the  given  circle.. 

Sue.    Prove  that  the  circle  described  on  DC 
is  the  required  locus. 


8.      What   has   been   done  to  the  proportion  —  =—    to  produce 

n      s 


s 


o+s      s 


9.     Construct  a  circle  having  a  given    radius    through  a  given 
point  and  tangent  to  a  given  line. 
SUG.     Find  two  loci  of  the  centers. 


PROPORTIONAL    LINES  137 

PKOPOSITION  XX. 

313.  THEOREM.  //  two  chords  intersect  the 
ratio  of  either  segment  of  the  one  to  either  seg- 
ment of  the  other  is  equal  to  the  ratio  of  the 
remaining  segment  of  the  second  to  the  remain- 
ing segment  of  the  first. 


Given  a  circle  with  chords  AB  and  CD  intersecting 
at  E. 

AE     CE        AE     DE 

To  Prove =  —  or  —  = 

DE     BE        CE     BE 

Proof.  SUG.  1.  As  no  A  are  given  in  the  theorem, 
two  A  must  be  constructed  one  of  which  contains 
the  required  antecedents  and  the  other  the  re- 
quired consequence.  Note  §  285  (1). 

2.  Prove     the     constructed    triangles 
similar. 

3.  Establish  the  required  proportions. 
Therefore — 

Make  a  second  construction  and  proof  for  this  proposi- 
tion. 

314.  COR.  The  product  of  the  segments  of  one  of 
two  intersecting  chords  of  a  circle  equals  the  product  of 
the  segments  of  the  other. 

To  Prove  AE*EB  =  CE*  ED. 

1.  The  line  joining  the  center  of  a  circle  with  an  outside  point 
bisects  the  angle  made  at  that  point  by  the  two  tangents  to  the 
circle. 


138  PLANE  GEOMETRY 

PBOPOSITION  XXI. 

315.  THEOREM.  //  two  secants  intersect  with- 
out a  circle,  the  ratio  of  the  first  to  the  second  is 
equal  to  the  ratio  of  the  external  segment  of  the 
second  to  the  external  segment  of  the  first. 


Given  a  circle  with  secants  AB  and  AC  intersecting 
the  circle  in  the  points  D  and  E  respectively. 

To  Prove  —  =±^ 

A  C     AD 

Proof.  The  desired  conclusions  will  at  once  follow 
if  two  triangles  can  be  constructed,  one  of  them  contain- 
ing the  required  antecedents  and  the  other  the  required 
consequents.  Try  such  a  construction  and  complete  the 
demonstration.  Note  §  285  (1). 

Therefore — 

316.  COR.  //  two  secants  meet  without  a  circle  the 
product  of  one  secant  and  its  external  segment  equals 
the  product  of  the  other  and  its  external  segment. 

1.  In  a  given  circle  two  radii  are  drawn    to  the   center   of   two 
chords,  and  the  angles  made  by  them   with   the  sect  joining    their 
feet  upon  the  chords  are  equal;  prove  the  choids  are  equal. 

2.  What  line  does  the  center  of  a    ladder  discribe  as  its  foot  is 
drawn  directly  away  from  the  building  against  which  it  leans,  the 
ground  beijig  level. 

3.  What  is  the  locus  of  the  center  of  a  given    sect  whose    end 
points  continually  touch  the  respective  side  of  a  right  angle'? 

Suo.     Ex,  2. 


PROPORTIONAL    LINES  139 

317.  THEOREM.  In  a  series  of  equal  ratios  the  ratio 
of  the  sum  of  the  antecedents  to  the  sum  of  the  conse- 
quents equals  any  of  the  given  ratios. 

-.         abed 
Given  -=--  =  -=-.  etc. 
a      b      c      d 

a  +6  +c  +cZ  +  .          a 
To  Prove  -    =— 

a'+6'+c'+d'+...     a' 

Proof.     SUG.     1.     Let  n  represent  the  common  value 
of  the  given  ratios.     Then  a  =  na',  b  =  nb',  etc. 

2.  .-.     a  -I-  b  +  c  +  d  +  .  .  .  = 
n  (af  +  V  +  c'+d'  +  ...) 

a  +  b  +  c  +  d  +  ...  a 

3.  .-.    =  M  =  -- 

a'+  b'+  c'+  d'+  ...  a' 

4.  Give   the   authority   for  each   step 
and  verify  by  a  particular  set  of  numbers. 

Therefore — 

1.  Give  a  summary  of  the  tests  for  similarity  of  triangles. 

2.  Through  a  given  point  draw  a  line  so  that  the  sect  in- 
tercepted between  two  given  intersecting  lines  is  bisected  at  the 
given  point. 

SUG.     Through  the  given  point,  o,  draw  a  line  parallel  to 
one  of  the  given  lines. 

3.  Through  a  point  included  be- 
tween the  sides  of  an  angle   draw  a 
line   so   that   the   sect  intercepted   be- 
tween   the   sides    shall    be    divided   in 
the  ratio  of  1  to  4. 

SUG.     Draw  MN\\AB.     Lay  off  on  side  AC  a  sect  MP 
equal  to  4AM.     Draw  PO  and  extend  to  AB. 
4         The  sum  of  the  three  perpendiculars  in  the  preceding  ex- 
ample equals  the  altitude  of  the  triangle. 

5.  The  three  altitudes  upon  the  three  sides  of  an  equilateral 
triangle  are  equal. 

6.  The   sum    of    the    perpendiculars    from    any    point    in    an 
equilateral  triangle  to  the  three  sides  is  constant. 


140  PLANE  GEOMETRY 

PROPOSITION  XXII. 

318.  THEOREM.    The   ratio   of  the  perimeters 
of  two  similar  polygons  equals  the  ratio  of  simil- 
itude of  the  polygons. 

Given  two  similar  polygons  P  and  P'  with  sides  a,  b,  c, 
. . .  and  a',  &',  c',  .  . .  respectively,  and  perimeters  p  and 
p'  respectively. 

To  Prove  £=*-• 

p'     a'- 

Proof.     SUG.     1.     Write  out  in  detail  the  relation  in- 
volving the  sides  of  P  and  P'. 

2.  Express   the  ratio  —    in  terms  of 

P' 
the  sides. 

3.  Complete  the  proof. 
Therefore— 

319.  MEAN  PROPORTION.     A  proportion  in  which  the 
means  are  the  same  is  a  mean  proportion. 

320.  MEAN  PROPORTIONAL.     The  second  or  third  term 
in  a  mean  proportion  is  a  mean  proportional. 

321.  THIRD  PROPORTIONAL.     The    fourth   term   of   a 
mean  proportion  is  the  third  proportional. 

a  _b 

^  —  ~    is  a  mean  proportion.     The  mean  proportional  is  b  and 

the  third  proportional  is  c. 

1.  Inscribe  in  a  circle  an  isosceles  triangle  with  a  diameter 
of  the  circle  as  its  base.     How  many  degrees  in  the  vertex  angle? 

2.  If  a  right  triangle  is  inscribed  in  a  circle,  the  hypothenuse 
is  always  a  diameter. 

3.  A  line  is  drawn  from  the  vertex  of  a  triangle  to    the   base. 
What  is  the  locus  of  a  point  that    divides  it    always    in  the  same 
ratio? 


PROPORTIONAL   LINES  141 

PEOPOSITION  XXIII. 

322.  THEOREM.     A  mean  proportional  of  two  num- 
bers equals  the  square  root  of  their  product. 

-.         a      b 
Given   -  =  — 
b      c 

To  Prove  b  =  V~ac. 

Proof.     ac  =  b'2.     Why?     Whence  6  =? 

4       x 
1.     I*  x=~9    find  *' 

ii.     li  3       x 

=  —  find  x. 

x       / 

3.     If  7-  =  7  find  a;. 

c>  iC 

PROPOSITION  XXIV. 

323.  THEOREM.     //  £/ie  altitude  on  the  hypot- 
enuse of  a  right  triangle  be  drawn.- 

I.  the  triangle  is  divided  into  two  triangles, 
each  similar  to  the  given  triangle  and  similar  to 
each  other; 

II.     the  altitude  is  a  mean  proportional  to  the 
segments  of  the  hypotenuse; 

III.  each  side  of  the  given  triangle  is  a  mean 
proportional  between  the  whole  hypotenuse  and 
the  adjacent  segment. 


A  MB 

Given  A  ABC  with  altitude  MC  drawn  to  the  hypote- 
nuse AB. 

To  Prove  I.  A  AMC  ^  A  A CB,  A  CMB  <-"  A  ACB, 
A  AMC  ""  A  CMB. 


142  PLANE  GEOMETRY 

II.     M C  a  mean  proportional  to  AM  and 

AM     MC 

MB,  i.  e. = 

MC     MB 

III.     AC  a  mean  proportional  to  AB  and 

AB     AC 

AM,  i.  e.  -  —  =  — — »  or  BC  a  mean  proportional  to  AB 
AC      AM 

AB     BC 

and  MB,  i.  e.  —  = 

BC     MB 

Proof.  I.  SUG.  L  A  is  common  to  &  ^L¥C  and 
ACS.  •'•  A  AMC"-*  A  AC£.  Why  ?  Complete 
the  proof  of  I. 

II.  SUG.  In  &AMC  and  CMJ5  sides  AM 
and  .¥ (7  are  homologous,  also  sides  MC  and  MB. 
Why?  Complete  the  proof. 

Apply  carefully  §  285  ( 1 )  note.  The  difficulty  in  this  case  lies  in 
the  fact  that  MC  is  in  two  different  triangles. 

III.     SUG.     Use  A  AMC  and  ACS,  noting  the 
equal   angles.      Select  the   homologous   sides   for 
each  ratio  in  accordance  with  the  suggestion  of 
§285  (1). 
Therefore— 

324.  COR.  In  a  semicircle  a  perpendicular  from  the 
arc  upon  the  diameter  is  a  mean  proportional  to  the  seg- 
ments of  the  diameter. 

NOTE:  In  reading  similar  polygons  care  should  be  observed  to 
follow  a  reading  which  gives  the  same  order  to  homologous  parts. 

1.  A  perpendicular  dropped  from  a  circle  upon  a  diameter  is 
a  mean  proportional  between  the  segments  of  the  diameter. 

2.  Use  the  preceding  exercise  to   construct  a  mean  propor- 
tional to  the  sects  a  and  b. 


EXERCISES 


143 


3.  By  means  of  this  proposition.  Til,  construct  a  mean  propor- 
tional to  two  given  sects.     Also  a  third  proportional. 

4.  If  the  altitude  upon  the  hypotenuse  divides  the  hypotenuse 
into  two  segments  of  4  and  6  in.  respectively,  what  is  the  length 
of  the  altitude?    What  is  the  length  of  each  side?     Express  results 
accurate  to  nearest  integer. 

One  segment  is  8  in.  and  the  adjacent  side  is  12  in.     Find  the 
hypotenuse  and  the  two  other  sides. 

5.  Of  all  parallelograms  having  equal  bases  and  altitudes  the 
rectangle  has  the  least  perimeter. 

6.  The  diagonals  of  an  isosceles  trapezoid  are  equal. 

7.  If  two  parallel  straight  lines  are  cut  by  a  transversal,  the 
bisectors  of  the  interior  angles  on  the  same  side  of  the  transversal 
are  perpendicular  to  each  other. 

8.  If  the  median  drawn  from  the  vertex  angle  of  a  triangle 
to   the   base   equals   half   the    base,    the   vertex    angle   is    a   right 
angle.     Of  which  exercise  is  this  the  converse? 

9.  If  a  side  of  a  triangle  is  parallel  to  the  bisector  of  an 
exterior  angle  of  the  triangle,  the  triangle  is  isosceles. 

10.  In  physics  one  learns  that  the  angle  of  incidence  equals 
the  angle  of  reflection.     In  the  figure  these  angles   are  i  and   r 
respectively.     Prove  that  an  object  viewed  in  the  mirror  M  seems 
as  far  behind  the  mirror  as  it  is  in  front  of  it. 

observer 


object 


11.  The   bisectors   of   the    angles    of   a    parallelogram    form   a 
rectangle. 

12.  a,  b,  c  are  the  angles  of  an  inscribed  triangle.     L  a  is  four 
times  b  and  b  is  one  seventh  of  c.     How  many  degrees  in  the  arcs 
of  the  circle  subtended  by  the  respective  sides  of  the  triangle? 

13.  In  the  figure  of  Prop.    XVI    draw  the  diameter  AM  and 
prove  the  theorem. 


144  PLANE  GEOMETRY 

PROPOSITION  XXV. 

325.     PROBLEM.      Construct    a    mean    propor- 
tional to  two  given  sects. 


Given  sects  a  and  b. 


a        T 
To  Construct  sect  x  so  that  -  =  — 


Construction.  SUG.  1.  Make  a  drawing  of  the  figure 
for  Case  II,  Prop.  XXIY  to  see  which  lines  must 
be  taken  for  a,  b,  x.  Then  construct  the  figure 
from  the  sects  a  and  b. 

2.     Or  in  a  similar  manner  make 
use  of  figure  and  conclusion  of  ex.  1  p.  142. 

1.        Construct  a  mean  proportional  to  two  given  sects  a  and  b 
by  use  of  Case  III  of  Prop.  XXIV. 

2..       The  bisectors  of  all  angles  inscribed  in  the  same  segment 
pass  through  a  common  point.     Where  is  that  point? 

3.  Draw    any   two   equal    non-intersecting   chords   of   a   circle 
and  connect  their  adjacent  extremities.  Prove  that  the  connecting 
chords   are  parallel. 

4.  Draw  any  two  parallel   chords  and  connect   their  extremi- 
ties.    Prove  that  the  connecting  chords  are  equal. 

5.  ABC  is  a  triangle  inscribed  in  a  circle  with  center  0.    Lino 
OD  is  perpendicular  to  BC.    Prove  L  DOC,  or  its  supplement,  equal 
to  ZA. 


PROPORTIONAL    LINES  145 

PBOPOSIT10N  XXVI 

326.  THEOREM.  //  from  a  common  point  a 
tangent  and  a  secant  be  drawn  to  a  circle,  the 
tangent  is  a  mean  proportional  between  the 
secant  and  its  external  segment. 


Given  a  tangent  AB  and  a  secant  AE  of  circle  C. 

„    _  AD     AB 

To  Prove  -  —  =  --- 

AB     AE 

Proof.  SUG.  1.  Indicate  a  triangle  having  AE  and 
AB  as  sides.  Do  the  same  for  AB  and  AD. 
These  4  are  similar.  Why? 

2.      Complete  the  demonstration. 
Therefore— 

327.  EXTREME  AND  MEAN  RATIO.  A  sect  is  divided 
into  extreme  and  mean  ratio  when  the  greater  segment 
is  a  mean  proportional  to  the  whole  sect  and  the  smaller 
segment. 

AB  is  divided  internally  at  M  in  extreme  and  mean    ratio  if 
AB       AM 
~A~M  =  llfB  an^   externally   at   M'   in   extreme   and  mean  ratio  if 


AM'  ~  M'B    B  A  w 

PEOPOSITION  XXVII. 

328.     PROBLEM.     To    divide    a    sect    internally 
and  externally  into  extreme  and  mean  ratio. 
Given  sect  AB. 

To  divide  AB  at  M  and  at  M1  so  that      —  =  —  —  and 


146 


PLANE  GEOMETRY 


AB      AM' 
AM'      M'B 


'~if  n  respectively. 


B  M  A  ~~M' 

SUG.  1.  Erect  a  perpendicular  at  B  (or  at  A) 
equal  to  |  AB.  With  the  free  extremity  0  of 
this  perpendicular  as  a  center  and  OB  as  a  radius 
describe  a  circle.  Connect  A  and  0  and  extend 
the  join  line  to  meet  the  circle  again  at  C.  Take 


2.  ^  =  ^  =  £L£:.      Why? 
AB     AD     AM 

3.  Form  a  new  proportion  from  thk  by 
division  and  complete  the  proof. 

4.  Therefore  M  is  the  required  point  of 
internal  division. 

II.     SUG.     1.     Extend  BA  to  M'  so  that  AM'=AC. 

2.  ^  =  ^.     Why? 
AC     AB 

3.  Form  a  new  proportion  from  this  by 
composition  and  complete  the  proof. 

4.     Therefore  M'  is  the  required  point  of 
external  division. 

329.  It  must  be  remembered  in  the  series  of  theorems  follow- 
ing that  in  the  operations  of  multiplication,  division,  squaring, 
etc.,  the  symbols  represent  numbers,  viz:  The  measures  of  the 
lines  of  the  same  name.  Sect  a  X  sect  &  means  the  product  of 
the  numerical  measures  of  sect  a  and  sect  &.  For  brevity  the 
operation  of  measuring  will  be  assumed  to  have  been  done  and 
the  name  or  notation  of  sect  will  also  represent  the  measure  of  it. 


NUMERICAL    RELATIONS  147 

The   operations  performed  in  the  demonstrations   are   then   those 
of  algebra.     §  252. 

1.  If  the  hypotenuse  is  divided  by  the  altitude  from  the 
right  angle,  the  ratio  of  the  squares  of  the  sides  equals  the  ratio 
of  the  adjacent  segments  of  the  hypotenuse. 

az       m 
To  PROVE     p  =  — •      (Use  fig.  of  Prop.  XXIV.) 

Sue.     a2  =  me.     Why?     b-  --/?" Complete  the  proof. 

PROPOSITION  XXVIII. 

330.  THEOREM.  The  square  of  the  hypote- 
nuse of  a  right  triangle  equals  the  sum  of  the 
squares  of  the  two  other  sides. 


Given  a  right  A  with  sides  a  and  b  and  hypotenuse  c. 

To  Prove  a2  +  62  =  c2. 

Proof.     Sue.     1.     a2  =  mc.     Why?    &2=? 

2.     a2  +  b2  =  ? 

Notice  that  m  and  n  are  here  coefficients  of  c. 
Therefore — 

Another   proof  is   found   in  §  376. 

331.  COR.     I.     The  square  of  either  side  of  a  right 
triangle  equals  the  difference  of  the  squares  of  the  hy- 
potenuse and  the  other  side. 

332.  COR.     II.     The  ratio  of  a  diagonal  TO  the  side  of 

v~2 
a  square  is  —  :=  V2 

2.  If  the  side  of  a  square  is  4  in.,  what  is  the  diagonal!    Find 
the  result  correct  to  two  decimal  places. 

3.  One  side  of  a  rectangle  is  4  in.  and  a  second  side  is  9  in. 
Find  the  diagonal  correct  to  two  decimal  places. 


148  PLANE  GEOMETRY 

4.  A  carpenter  squares  the  frame  of  a  building  by  measur- 
ing 6  ft.  from  the  corner  on  one  side,  8  ft.  from  the  corner  on  the 
other,  and  then  draws  the  frame  until   the   distance  across  from 
the  ends  of  the  sects  is  10  ft.     Verify  the  correctness  of  his  plan. 

5.  The  distance  from  A  to  B  is  desired.     A  distance  70  rods 
is  measured  from  B  to  C,  so  that  the  angle  BAG  is  90°.     If  AC 
is  found  to  be  60  rods,  how  long  is  AB  ? 

6.  The  diagonal  of  a  rectangle  is   12  in.     Find  the  sum  of 
the  squares  of  the  four  sides. 

7.  Two  sides  of  a  rectangle  are  6  and  8.     Find  the  diagonal. 
Do  the  same  for  sides  5  and  12;  6  and  11. 

8.  Construct  a  right  triangle  and  the  altitude  upon  the  hy- 
potenuse.    Measure  the  segments  of  the  hypotenuse  and  compute 
the  altitude.     Check  your  work  by  measuring  the  altitude. 

0.        Find  a  mean  proportional  to  3  and  5;   to  4  and  7;   to  G 
and  11.     Use  algebra  and  also  geometric  construction. 

10.  If  #  =  V   3  X  4   find  x.     Sug.     x  is  a  mean  proportional 
to    3    and   4.     Why?      Construct   x    geometrically    and    verify   by 
measurement. 

11.  Find  a  line  the  length  of  which  is  V~12~. 

SUG.     VTU  =  V3  x  4.     Make  two  other  constructions.    Use 
§  324. 

12.  Find    v/~8~  by  the  measurement  of  a  line.  Also  V~57     VT^ 
Vl5. 

Sue.     Use    §  324  and    verify   results  by  extracting   the 
roots. 

13.  A  boy  in  traveling  from  his  home  A  to  a  town  B  on  his 
wheel  traveled  16  mi.  in  a  straight  line  and  then  24  mi.  in  a  line 
at  right  angles  to  the  first  road.     How  many  miles  would  he  have 
saved  by  traveling  along  the  railroad  which  ran  in  a  straight  line 
from  A  to  U? 

14.  The  sum  of  the  squares  of  the  two  diagonals  of  a  rectangle 
equals  the  sum  of  the  squares  on  the  four  sides. 

15.  Find  the  length  of  the   side  of  a  square   correct   to   two 
decimal  places  if  the  diagonal  is  1.     If  the  diagonal  is  18;  2;  32; 
26;  4;  12. 

16.  Find  the  diagonal  of  a  square  the  side  of  which  is  8.     Do 
the  same  if  the  side  is  25;  3.15;  4.18. 


NUMERICAL    RELATIONS  149 

PROPOSITION  XXIX. 

333.  THEOREM.    The  square  of  the  sum  of  two 
sects  equals  the  sum  of  the  squares  of  the  sects 
plus  twice  the  product  of  the  sects. 

Sue.  If  a  and  b  are  the  measures  of  the  two 
given  sects,  §  329,  it  is  necessary  and  sufficient  to 
prove  the  formula  (a  +  b)2  =  a2  +  b2  +  2ab. 

The    pupil   should   do   this  by   performing   the   indicated   multi- 
plication. 

Another  proof  is  found  in  P.  186, 

PROPOSITION  XXX. 

334.  THEOREM.     The  square  of  the  difference 
of  two  sects  equals  the  sum  of  the  squares  of  the 
sects  minus  twice  the  product  of  the  sects. 

Given  a  and  b  as  the  measures  of  the  two  sects. 
To  Prove   (a  —  b)2  =  a2  +  b- —  2  ab. 
Proof  left  to  the  pupil. 

335.  PROJECTION  OF  A  POINT.   The  A 
projection  of  a  point  upon  a  line  is 

the  foot  of  the  perpendicular  drawn 
from  the  point  to  the  line.  3 

If  AM  is  perpendicular  to  EC,  then  M  is  the  projec- 
tion of  A  upon  BC. 

336.  PROJECTION  OF  A   SECT.     The  projection   of   a 
straight  sect  upon  a  straight  line  is  the  sect  between  the 
projections  of  the  extremities  of  the  given  sect  upon  the 
same  line. 

1.        The  product  of  the  sum  of  two  sects  by  their  difference 
equals  the  difference  of  the  squares  of  the  sects. 

To  PROVE  (a -I- 6)    (a — 6)  =  a?  —  £2. 


M 


150 


PLANE  GEOMETRY 


PROPOSITION  XXXI. 

337.  THEOREM.  The  square  of  the  side  oppo- 
site an  acute  angle  of  a  triangle  equals  the  sum  of 
the  squares  of  the  two  other  sides  minus^the 
product  of  one  of  those  sides  and  the  projection 
of  the  other  upon  it. 


Given  a  A  with  the  sides  a,  b,  c,  side  a  being  opposite 
an  acute  angle  and  m  being  the  projection  of  b  upon  c. 

To  Prove  a2  =  b2  +  c2  —  2mc. 

Proof.     SUG.     1.     Drop  a  perpendicular,  p,  from  ver- 
tex C  to  side  c. 

2.     a2=p2  +w2  =  p2+  (c-m)*  = 
&2  -  m2  +  (c  -  m) 2  =  b2  -  m2  +  32  +  m2  —  2wir  == 
62  +  c2  — 2wc. 

Therefore — 

1.  If   a    quadrilateral   is   circumscribed   about    a   circle,    the 
sum    of    one    pair    of    opposite  sides  is  equal  to  the  sum  of  the 
other  pair. 

2.  If    two    circles    are    tangent    and    two  secants  are   drawn 
through   the    point   of   contact  the   two 

chords  joining  the  points  in  which  the 
secants  meet  the  respective  circles  are 
parallel. 

PROVE  AD  |  BC. 

Sue.     Draw  the  common  tangent 
MN.     Compare  &  AOM  and  CON  ; 
A  AOM  and    ADO;  A. 
OBC. 


NUMERICAL    RELATIONS 


151 


PROPOSITION  XXXII. 

338.  THEOREM.  The  square  of  the  side  oppo- 
site an  obtuse  angle  of  a  triangle  equals  the  sum 
of  the  squares  of  the  two  other  sides  plus  twice 
the  product  of  one  of  those  sides  and  the  projec- 
tion of  the  other  upon  it. 


Given  a  A  with  sides  a,  b,  c,  side  a  being  opposite  an 
obtuse  angle,  and  m  being  the  projection  of  b  upon  c. 

To  Prove  a2  =  b2  +  c2  +  2mc. 

Proof.     Suo.     1.     The  projection  m  will  in  this  case 
lie  on  the  extension  of  c. 

2.     Complete   the   proof   in   a   manner 
similar  to  the  method  of  §  337. 

339.  The  theorems  of  §  337  and  §  338  may  be  in- 
cluded in  one  statement  which  may  be  proved  by  the 
principle  of  continuity  (§  310)  as  soon  as  a  distinction  is 
made  as  to  the  direction  of  the  line  representing  the 
projected  side.  The  theorem  may  be  stated  in  general 
as  follows : 

The  square  of  any  side  of  a  triangle  equals  the  sum  of 
the  squares  of  the  two  other  sides  minus  twice  the 
product  of  one  of  those  sides  and  the  projection  of  the 
other  upon  it.  c  c- 


152  PLANE  GEOMETRY 

Given  a  A  with  sides  a,  b,c,  LA  being  acute  or 
obtuse,  or  right,  and  m,  the  projection  of  h  upon  c,  be- 
ing considered  as  positive  when  it  extends  in  the  direc- 
tion AB  and  negative  when  it  extends  in  the  opposite 
direction. 

To  Prove  a2  ~=  62  +  c2  —  2wc. 

Discussion.  As  angle  A  increases  from  an  acute  an- 
gle to  an  obtuse  angle,  passing  through  the  value 
90°,  point  M,  the  projection  of  vertex  C  upon  side  c 
moves  towards  the  vertex  A,  passes  through  this  point 
when  Z  A  —  90°,  and  passes  out  on  the  extension  of 
side  c  through  A  when  Z  A  becomes  obtuse.  For  Z  A 
acute,  m  is  positive ;  for  Z  A  =  90°,  m  is  0;  and  for  Z  A 
obtuse,  m  is  negative. 

The  formula  a2  =  62  +  c2  — 2wc  for  these  three 
values  of  m  reduces  to  the  formulas  previously  proved 
for  these  three  cases  in  §§  337,  330,  and  338  respectively. 

This  theorem  will  be  met  again  in  trigonometry  under 
the  name  of  The  Law  of  Cosines. 

1.  Fasten  a  rubber  band  to  the  two  points  of  a  pair  of  di- 
viders.    Let  the  rubber  band  represent  side  a  in  the  triangles  of 
§  339,  and  the  arms  cf  the  dividers  the  sides  b  and  c.     The  angle 
made  by  the  arms  then  represents  angle  A.     As  the  dividers  are 
slowly  opened,  note  the  change  of  LA  from  acute  values  to  90° 
and   then    to    obtuse   values   and    at    the    same   time    observe    the 
shortening  of  the  projection  of  &  on  c  from  a  positive  sect  through 
zero  on  to  increasingly  large  negative  sects. 

2.  The  sum  of  the  squares  of  the  diagonals  of  a  parallelogram 
equals  the  sum  of  the  squares  of  the  four  sides. 

3.  The  sum  of  the  squares  of  the  diagonals  of  a  quadrilateral 
equals  the  sum  of   the  squares  of  the   four    sides   plus  four  times 
the  square  of  the  sect  joining  the  mid-points  of  the  two  diagonals. 

4.  A  chord  of  a  circle,  AB,  is  5  in.  long.     If  it  be  produced 
to  a  point  C  so  that  the  external  segment  BC  is  10  in.,  how  long 
is  the  tangent  from  C? 


NUMERICAL    RELATIONS 


153 


PROPOSITION  XXXIII. 

340.     PROBLEM.     To  find  any  altitude  of  a  tri- 
angle in  terms  of  its  sides. 


Given  A  with  sides  a,  b,  c,  and  ha  the  altitude  upon  a. 
To  find  ha  in  terms  of  a,  b,  c. 

SUG.    1.  ha*  =  &2—  ZJC2  and  c2  —  a2+62—  2aX/>C. 
2.     .  •  .     DC  = 


2a 


3. 


, 


(c+a-b)  (a+b-c)  (a+b+c) 


4.     Let  2*  = 
2(s— a)  —  c  —  a  +  &,  2(s— b)  —  c+a— b, 
2(s—c)  =  a  +  b— c,  whence, 

2 ^ 


2    -^- 


5.      Similarly  /< ,  =  —  V  (s—a)(s—b)(s—c)s 
b 

2 

AV  =  -  V (s—a)(s—b)(s—c)s 
c 

1.     By  formula,  find  the  three  altitudes  of  triangle  the  sides 
of  which  are  12,  14,  18.     Construct  the  triangle  and  measure   the 


154  PLANE  GEOMETRY 

altitudes  as  a  check  on  the  computations.     Do   the  same  for  the 
triangle  with  sides  7,  9,  and  12;  also  13,  15,  and  20. 

PROPOSITION  XXXIV. 

341.  THEOREM.  In  any  triangle  the  product 
of  any  two  cides  equals  the  product  of  the  alti- 
tude upon  the  third  side  and  the  diameter  of  the 
circumscribed  circle. 


I 

Given  A  ABC  with  sides  a,  b,  c,  altitude  ha  upon  side 
a  and  AM  or  d  the  diameter  of  the  circumscribed  circle. 
To  Prove  bc  =  dha. 

Proof.     SUG.     1.     Connect   M   and  B.     A  CAD   and 
MAB  are  similar.     Why? 

2.     Deduce   a   proportion   from  which 
the  required  equality  may  be  obtained. 
Therefore— 

PROPOSITION  XXXV. 

342.  PROBLEM.  To  find  the  length  of  the 
radius  of  the  circumscribed  circle  in  terms  of  the 
sides  of  the  triangle. 

Given  A  ABC  with  sides  a,  b,  c,  altitude  ha  upon  side 
a,  and  AM  or  d  the  diameter  of  the  circumscribed  circle. 
To  find  d  in  terms  of  a,  ft,  c.  (See  fig.  of  Prop.  XXXIV,) 
Sue.     1.     haxd  =  bc.    Why? 


NUMERICAL    RELATIONS  155 

3.  Substitute  for  ka  its  value  in  terms  of 
a,  b,  c. 

4.  Simplify  this  result,  obtaining 

abc 


4Vs(s— a)  (s—b)  (s—c) 

1.  Find  the  radius  of  the  circle  circumscribed  about  a  tri- 
angle with  sides  8,  9,  and  12.  Construct  the  figure  and  check 
the  computation  by  measurement. 

PROPOSITION  XXXVI. 

343.  THEOREM.  In  any  triangle  the  square  of 
the  bisector  of  any  angle  equals  the  product  of 
the  sides  including  the  angle  minus  the  product  of 
the  segments  of  the  third  side  made  by  the  bisec- 
tor. 


c 

Given  A  ABC  with  BD  bisecting  L  B,  and  meeting 
AC  at  D. 

To  Prove  ~BD*  =  AB  x  BC  —  AD  X  DC. 
Proof.     SUG.     1.     Circumscribe  a  circle  and  extend 
BD  to  meet  the  circle  at  M.     Draw  MC  (or  MA). 
2.     &ABD^&MBC.     Why? 


3.  =  Why? 

MB     BC 


4.     . 

BD 


5.  .'.  BD2  =  AB  x  BC  —  AD  X  DC. 

6.  Verify  each  step. 
Therefore  — 


156  PLANE  GEOMETRY 

PKOPOSITION  XXXVII 

344.  PROBLEM.  To  find  the  length  of  the  bi- 
sector of  an  angle  of  a  triangle  in  terms  of  the 
sides  of  the  triangle. 


D    a 

Given  A  ABC  with  sides  a,  b,  c,  AD  or  dn    being  the 
bisector  of  Z  A,  meeting  EC  at  point  D. 

To  find  an  expression  for  da  in  terms  of  a,  &,  c. 

SUG.    1.      By  §  343  da*  =  bc- BDxDC.     Also 
##__c 
Z)6'~6      Why? 

c          &  c+&          c+& 

r>/?  n  ]\ 

3. 


b+c  b+c 

ac       ab  a*bc 

=  ?).^  —      —  x  --  =  be— 

b+c     b+c  (b+c)* 

c  _  be  \_(b+c)*  —  a2  ]_ 


_bc(b+c—a)  (a+b+c) 


(b+c)2  (b+c) 


2 


_ 

da  =  -    -  Vbcsfb—a),   in    which 


2.s  =  a+6+c,  as  in  §  340. 

6.     Verify   each   step   and   write    corre- 
sponding expressions  fcr  dfi  and  r/  . 


NUMERICAL    RELATIONS  157 

1.  If  the  sides  of  a  triangle  are  7,  9,  and  12;  find  the  length 
of  the  bisector  of  each  angle.     Construct  the  triangle  and  check 
the  computations  by  measurements. 

2.  In   triangle   ABC   side   a  =  24,    &  =  17,    and    c  =  20;    find 
the  bisector  of  L  C. 

3.  Find  the  radius   of  the  circle  circumscribed  about  a  tri- 
angle with  sides   of   12,   18,  and   20.     Check  the   results  by   con- 
struction. 

PROPOSITION  XXXVIII. 
345.  THEOREM.  In  any  triangle  if  a  median 
be  drawn  to  the  base  the  sum  of  the  squares  of 
the  two  other  sides  is  equal  to  twice  the  square 
of  half  the  base  plus  twice  the  square  of  the 
median. 


D 

Given  a  A  with  sides  a,  b,  c,  with  D  the  projection  of 
C  and  d  the  projection  of  mc  on  c,  and  mc  the  median  on 
side  c. 


To  Prove  a2  +  62  =  2m2  +  2 

'U 

Proof.  SUG.  1.  If  a  and  b  are  unequal,  one  angle 
between  mc  and  c  is  acute  and  the  other  is 
obtuse.  Suppose  then  that  a  is  opposite  an 
obtuse  angle. 

2.  Express  a2  and  b2  in  terms  of  the 
remaining  sides  of  the  two  triangles  made  by 
mc.     Add  the  resulting  expressions. 

3.  Prove  the  proposition  when  sides  a 
and  b  are  equal. 

Therefore — 


158  PLANE  GEOMETRY 

1.  Prove  §   343   for  the  bisector  of 
an  exterior  angle. 

Sue.  Letter  the  figure  as  in  the 
original  proposition.  Notice  that 
AC  is  divided  externally  at  D  and 
AC  =  AD  —  CD. 

2.  Discuss  the  above  problem  when  the 
angle    bisected    is    exterior    to    the    vertical 
angle  of  an  isosceles  triangle.     In  this  case 
BD  is  parallel  to  AC.    Why?    No  conclusions 
can  be  drawn  as  in  the  other  cases.  BD,  AD, 
CD,  all  become  infinitely  large. 

PROPOSITION  XXXIX. 

346.  PEOBLEM.  To  find  the  length  of  any  me- 
dian in  terms  of  the  sides. 

Given  A  ABC  with  sides  a,  b,  c  and  ma  the  median 
on  side  a. 

To  express  ma  in  terms  of  a,  b,  c. 

From2w?  =  b*+c*-2l  -  \«  and 


Write  the  corresponding  values  of  rinb  and  mc. 
*This  proposition  may  be  omitted. 

PROPOSITION  XL. 

347.  THEOREM.  In  any  triangle,  if  the  median 
be  drawn  to  the  base,  the  difference  of  the  squares 
of  the  two  sides  equals  twice  the  product  of  the 
base  and  the  projection  of  the  median  on  the  base 

Given  the  conditions  of  Prop.  XXXVIII. 
To  Prove  a2  —  b2  =  2cd. 
Proof.     Subtract  b2  from  a2. 
Therefore — 


EXERCISES  159 

1.  In    &ABC    side    a  =  8,    &  =  12,    and    c  =  14.      Find    the 
length  of  the  median  on   c.     Construct   the   triangle   and  median 
and  cheek  the  computations  by  measurement. 

2.  In  the  above  example  find  m^  and  m     and  check  computa- 
tions. 

Given  the  sides  a  =  10,  &  —  6,  and  c=8.  Find  the  median  on 
a  and  check  the  computation. 

3.  Draw  a  triangle  and  a  median  with  a  straight  edge.   Meas- 
ure the  sides  and  the  median.     Check   the   results   by  computing 
the  length  of  the  median,  and  also  by  using  other  theorems  in- 
volving the  median  and  sides. 

4.  If  two   circles  are  tangent  and  a  sect  be  drawn  through 
the  point  of  tangency  terminated  by  the  circles. 

I.     Tangents  to  the  circles  at  the  extremities  of  the  sect  are 
parallel. 

SUG.  Draw  the  common  tangent  at  the  point  of  tangency. 
Consider  the  two   cases  of  internal  and  external  tangency. 
II.     The  diameters  of  the  two  circles  through  the  extremities 
of  the  sect  are  parallel. 

Sue.  Use    same     construction    as  above,  with  two  cases. 

5.  A  ABC    is    equilateral    and   AB    and 
AC  are  tangents  to   the  circle.     How  many 
degrees  in  arc  BC1     In  arc  BMC1  M( 

6.  If  AB  is  18  in.,  what  is  the  diam- 
eter of  the  circle? 

7.  Show  that  OB  is  a  mean  proportional  to  OE  and  OA. 

OE  _  OB 
To  PROVE  Q2  -  02- 

318.  EXTERNAL  TANGENT.  A  line  is  an  external  tan- 
gent to  two  circles  if  it  is  tangent  to  each  but  does  not 
cut  the  line  of  centers. 

INTERNAL  TANGENT.  A  line  is 
an  internal  tangent  to  two  circles 
if  it  is  tangent  to  each  and  cuts 
the  line  of  centers. 

AB  is  an  external  tangent  and  CD  an  internal  tan- 
gent. 


160  PLANE  GEOMETRY 

PROPOSITION  XLI. 

349.     PKOBLEM.     To    construct   a   common   ex- 
ternal tangent  to  two  circles. 


Given  two  ©  C  and  C'. 

To  Construct  AB  a  common  external  tangent  to  ©  C 
r.nd  C'. 

SUG.  1.     Construct    a    circle    concentric    with 
the   larger   of   the    given   circles    with    a    radius 
equal  to  the  difference  between  the  given  radii. 
Let  this  radius  be  CM.    Why  ? 

2.  Draw  a  tangent  to  this  auxiliary 
circle  from  C' '.    How  may  this  be  done? 

3.  Draw   perpendiculars  to  this  tan- 
gent line  from  C  and  C'  meeting  the  circles  in 
the  points  A  and  B  respectively.    Why  ? 

4.  AB  is  the  required  external  tan- 
gent.    Explain  why. 

5.  Is  AB  the  only  external  tangent? 
In  what  step  of  the  construction  might  a  second 

one  be  introduced? 

1.  The  common  tangent  to  the  two  ex- 
ternally    tangent     circles     at     the     common 
point  meets  an  external  tangent  in  a  point 
equally  distant  from  the  three    points    of 
contact.     To  prove  that  OA  =  OB  =  OC. 

2.  If  two  circles  are  tangent  externally  the  sects  which  join 
the  point  of  contact  to  the  two  contact  points  of  an  external  1;m- 
gent  form  a  right  angle.     To  prove  L  ACB  =90°. 


PROBLEMS    OP    CONSTRUCTION  161 

1.  Two  parallel   tangents   to   a  circle  intercept   a  sect   on   a 
third  tangent.     Prove  that  this  sect  subtends  an  angle  of  90°   at 
the  center. 

2.  If  two  circles  are  externally  tangent 
the  sect  which  is  the  common  external  tan- 
gent is  a  mean  proportional  to  the  diameters. 

3.  Construct    a    mean    proportional   to 
sects     a    and     b     by    means    of  Ex.  2. 

PROPOSITION  XLII. 

350.     PROBLEM.     To  construct  an  internal  tan- 
gent to  two  circles. 


Given  ®  C  and  C". 

To  Construct  AB,  an  internal  tangent  to  ©  C  and  C' . 

SUG.     1.     Draw   a  circle   concentric   with   the 

smaller  of  the  given  circles  and  with  a  radius 

equal  to  the  sum  of  the  radii  of  the  given  circles. 

Why? 

2.     The     completion    of    the     proof    is 
left  to  the  pupil. 

351.  In  the  previous  pages,  certain  principles  of  con- 
struction were  stated  and  certain  constructions  made. 
In  most  of  the  problems  thus  far  the  pupil  has  been 
given  more  or  less  aid.  In  order  that  the  pupil  may 
become  more  independent  in  the  solution  of  problems 
certain  methods  of  analysis  will  now  be  studied  in  con- 
nection with  the  solution  of  a  few  problems. 


162  PLANE  GEOMETRY 

Constructions  are  based  upon  previously  demon- 
strated theorems,  and  the  difficult  thing  in  the  solution 
of  any  problem  is  the  recognition  of  the  previous  propo- 
sitions which  may  be  applied  in  effecting  the  desired 
construction.  In  the  more  difficult  problems  this  can 
most  easily  be  done  by  first  making  a  sketch  of  the  com- 
pleted construction.  This  construction  should  next  be 
analyzed  in  order  to  discover  the  theorems  involved; 
then  by  reversing  the  order  of  the  analysis,  the  con- 
struction can  usually  be  made. 

Each  problem  usually  gives  certain  conditions  for 
the  determination  of  one  or  more  points,  the  geometric 
representation  of  which  conditions  are  loci,  their  in- 
tersections being  the  required  points.  The  number  of 
solutions,  when  a  problem  calls  for  the  location  of  a 
point,  is  the  same  as  the  number  of  such  intersections. 
In  the  case  of  no  possible  intersection,  there  is  no  solu- 
tion. 

352.  PROB.  I.  To  construct  a  triangle,  given  one 
side,  an  adjacent  angle,  and  the  altitude  upon  that  side. 

Given  side  AB,  L  A,  and  alti- 
tude MC. 

As  AB  is  given  it  can  be  drawn 
and  at  one  extremity  the  given  an- 
gle A  can  be  constructed. 

The  vertex  C  must  now  be  determined.  One  locus  of 
C  is  the  unlimited  side  of  L  A,  i.e.  line  AC.  The  other 
locus  of  C  must  be  found  from  the  other  condition,  viz : 
the  given  altitude  MC.  It  is  evident  that  the  locus  of 
the  vertex  of  a  triangle  with  a  given  altitude  is  a  line 
parallel  to  the  base  and  at  a  distance  from  the  base  equal 
to  the  given  altitude.  The  construction  can  now  bo 

easily  made. 


PROBLEMS    OF    CONSTRUCTION  163 

Discussion:  On  a  given  side  of  line  AB  there  is  but 
one  solution  for  two  straight  lines  can  have  but  one  in- 
tersection. Also  there  is  always  a  solution  for  if  a  line 
cuts  one  of  two  parallels  it  will  cut  the  other. 

PROB.  II.  Upon  a  given  sect  to  construct  a  segment 
of  a  circle  which  shall  contain  a  given  angle. 

Given  Z  n  and  sect  BC. 
The  problem  is  easily  solved 
if  the  center  of  the  required 
circle  is  found.  c- 

One  locus  of  the  center  is 
the  perpendicular  bisector  of  sect  BC.  Why  ?  it  is  ob- 
served from  the  figure  that  if  the  segment  B  A  C  con- 
tains Z  A  =  L  n  that  the  arc  BC  is  not  only  twice  the 
measure  of  LA  but  also  of  the  angle  at  B  between  CB 
and  a  tangent  to  the  circle.  In  other  words  if  an  angle 
equal  to  Ln  be  constructed  at  B,  the  second  side  will  be 
tangent  to  the  required  circle.  The  perpendicular  to 
this  tangent  at  B  is  also  a  locus  of  the  center.  The  cen- 
ter is  thus  determined.  Does  this  problem  have  a  solu- 
tion for  every  sect  BC  and  Z  n  ? 

PROS.  III.  To  construct  a  circle  with  a  given  radius 
that  shall  pass  through  a  given  point  and  be  tangent  to  a 
given  circle. 

Given  a  circle  C,  a  fixed 
point  P,  and  a  sect  r. 

To  Construct  a  circle  with 
radius  r,  passing  through  P, 

and  tangent  to  C. 

Draw  a  figure  representing 
the  completed  construction. 
The  center  C'  of  the  required 
circle  is  needed.  What  is  the 


164  PLANE  GEOMETRY 

distance  from  C'  to  Cf  What  then  is  a  locus  of  G'? 
Since  the  distance  PC'  is  r,  what  is  a  second  locus  of  C'  ? 
Discussion:  In.  general  how  many  intersections  of 
these  two  loci  and  hence  how  many  solutions  ?  Is  it  pos- 
sible for  one  solution  only  to  exist?  In  this  case  how 
would  the  distance  PC  compare  with  the  radii  of  the  two 
circles?  What  relation  of  distances  would  make  the  in- 
tersection of  the  two  loci  an  impossibility?  Illustrate 
these  special  cases  by  figures. 

PROB.  IV.  To  construct  a  triangle  having  given  the 
base,  the  altitude,  and  the  radius  of  the  circumscribed 
circle.  ^ 

To  Construct  a  triangle  ABC  with        b 

the  sects  a,  ~b,  and  c  as  base,  altitude,    « — _ 

and  radius  of  the  circumscribed  circle  respectively. 

The  problem  is  solved  as  soon  as  one  finds  the  third 
vertex,  A,  the  extremities  B  and  C  of  sect  a  being 
two  of  them. 

What  is  the  locus  of  vertex  A  as  determined  from  the 
given  altitude?  The  center  of  the  circumscribed  circle 
is  at  distance  r  from  B  and  C.  Locate  its  center  0. 
The  circumscribed  circle  is  the  second  locus  of  vertex  A. 

Discussion.  The  first  condition  determines  as  a  lo- 
cus for  A  two  straight  lines. 
Why  two?  The  Second 
condition  determines  two  cir- 
cles. Why?  If  the  altitude 
is  less  than  MD,  each  of  the 
straight  line  loci  meets  each 
circle  locus  twice.  How 
many  solutions?  If  the  al- 
titude is  greater  than  MD 


EXERCISES  165 

but  less  than  MP,  each  straight  line  locus  meets  one 
circle  locus  twice.  How  many  solutions?  If  the  al- 
titude equals  MD  and  is  less  than  MP  each  straight  line 
locus  is  tangent  to  one  circle  locus  and  meets  the  other 
in  two  points.  How  many  solutions?  If  the  altitude 
equals  MP  each  straight  line  locus  is  tangent  to  one  cir- 
cle and  does  not  meet  the  other.  How  many  solutions? 
If  the  radius  equals  one  half  the  base,  then  MD  =  MP, 
the  problem  is  impossible  or  there  would  be  two  or  four 
solutions,  and  any  of  the  above  cases  in  which  MD  and 
MP  are  assumed  unequal  are  impossible.  Discuss  this 
in  detail. 

If  the  radius  is  less  than  half  the  base,  there  is  no 
solution.     Why  ? 

1.  The  locus  of  the  vertex  of  a  given  angle  subtending  a  given 
sect  is  the  arc  of  the  segment  which  contains  the  angle. 

SUG.    1.     Construct    the    segment    containing    the    given 
angle. 

2.  All  angles  inscribed   in  this  segment  equal  the 
given  angle.     Why? 

3.  Any    angle   not    inscribed   in   this   segment   and 
subtending  the  given  sect  is  either  greater  or  less  than  the 
given  angle.     Why! 

2.  Divide  a  sect  8  in.  long  internally  into  extreme  and  mean 
ratio.      Measure   the   segments   and   arithmetically   check   the   con- 
struction. 

3.  Divide   8   into   two   parts   such   that   one  part  is   a   mean 
proportional  between  8  and  the  other  part. 

4.  Given  a  chord  in  a  circle.     From  any  point  in  the  arc  draw 
a  second  chord  which  shall  be  bisected  by  the  first.    Two  solutions. 


166  PLANE  GEOMETRY 

5.        Construct  a  triangle,  given  the  base,  the  altitnde,  and  the 
vertex  angle. 

0.        Construct  a  triangle  given  two  angles  and  a  side  opposite 
one. 

7.  Construct   a   triangle,   given   two    angles    and   the   altitude 
upon  the  included  side. 

8.  Construct   a   triangle,   given   the   base,   the   median   to   the 
base,  and  an  angle  adjacent  to  the  base. 

9.  Construct   a  triangle,   given   the   base,   the  median   to   the 
base,  and  the  vertex  angle. 

10.  Construct   a   circle    of    given    radius   and    tangent    to    two 
given  circles.     What  conditions  are  necessary  for  a  solution?    How 
many  solutions? 

11.  Construct   a    circle    having   a   given    radius,    tangent    to    a 
given  line,  and  passing  through  a  given  point.     How  many  solu- 
tions and  what  conditions  govern  each  case? 

12.  Construct  a  mean  proportional  to  two      sects,  using  Ex.    1 
p.  142. 

13.  Construct  a  third  proportional  to  two  sects. 
GIVEN  sects  in  and  n. 

m  _   n 
To  construct  sect  x  such  that  ~       — • 

It  Ju 

SUG.     1.     On     an  un- 
limited   line    lay    off    a  m 
sect      m  +  n.        Through  m 

one  extremity  of  this  sect  draw  an  oblique  line  and  from 
the  intersection  lay  off  on  the  line  sect  n.  Join  the  ex- 
tremities of  the  two  sects  of  lengths  m  and  n  and  through 
the  free  extremity  of  sect  m  +  n  draw  a  line  parallel  to  this 
join  line.  The  sect  intercepted  on  the  oblique  line  by  the 
parallels  is  x. 

2.     Verify   the   construction.  ft 

14.  Construct    a    fourth    proportional   to  f 

sects  a,  I),  c,  using  the  method  of   Ex.    13. 


EXERCISES  167 

15.  Construct  a  fourth  proportional  to   three  sects  using  the 
problem  for  constructing  a  triangle  similar  to  a  given  triangle. 

16.  Construct  a  fourth  proportional  to  three  sects  using  Ex. 
1,  p.  120. 

NOTE:     For  such  a  problem  any  theorem  in  which  a  propor- 
tion of  four  different  terms  has  been  established  may  be  used. 

17.  If  a  chord  is  12   in.  long  and  the  perpendicular  to   that 
chord  bisecting  the  subtended  arc  is  3  in.,  find  the  radius. 

18.  Draw  accurately  the  figure  of  the  preceding  exercise,  con- 
struct the  circle  and  check  the  computations  by  measurements. 

19.  A   bridge  with  a   circular  stone   arch   has  a   39    ft.   span 
and  an  8  ft.  rise.    What  is  the  diameter  of  the  circle? 

20.  The  altitude  of  an  equilateral  triangle  is  12  in.     Find  the 
length  of  a  side. 

21.  Prove  that  the  perpendiculars  from  the  center  of  a  circle 
to  the  sides  of  an  inscribed  equilateral  polygon  are  equal.     This 
perpendicular  is  the  apothem  of  the  polygon. 

22.  Circumscribe   a    circle   about    an   equilateral   triangle   and 
compare  the  apothem  with  the  radius  and  the  altitude. 

23.  Given    an   equilateral   triangle   of   side    12   in.     Find    the 
apothem  and  altitude.    Do  the  same  for  a  triangle  of  side  a. 

24.  Connect  the  vertices   of  an  inscribed  equilateral  triangle 
with  the  mid-points  of  the  arcs  between   them.     Prove   that  the 
resulting  hexagon  is  equilateral  and  equiangular. 

25.  Prove  that  the  side  of  an  inscribed  equilateral  hexagon 
equals  the  radius. 

26.  The  sides  of  a  triangle  are  8,  11,  13.     Find  the  altitudes, 
medians,   bisectors   of  the  angles,  and  the  radius  of  the  circum- 
scribed circle. 

27.  In  a  circle  of  10  in.  radius  there  are  two  parallel  chords, 
one  being  6  in.  from  the  center  and  the  other  8  in.     Find  their 
lengths. 

28.  Find  a  fourth  proportional  to  2,   7,  and  8  by  arithmetic 
and  geometric  methods.     Check  the  results  by  comparison. 

29.  Find  a  fourth  proportional  to  5,    7,  and    9  by  arithmetic 
method    and    by    geometric    construction.      Check    the    results    by 
comparison. 

30.  Find  by  arithmetic  method  and  by  geometric  construction 
a  sect  equal  to  V~9;  VT;  VTO.     Compare  the  results. 


1f>8 


PLANE  GEOMETRY 


31.  If  two   circles  intersect,  tangents   drawn   from  any  point 
in  their  common  chord  extended,  are  equal. 

32.  CD  is  tangent  to  circle  0,  AB  is  a 
secant  cutting  the  circle  at  points  A  an  B 
and    perpendicular   to   CD.     At   what   point 

X  in  CD  is  L  AXB  the  greatest  ?  E 

33.  If  A  and  B  are  goal  posts  on  a  fcot 
ball  field  and  E  the  place  of  a  touch  down, 
how  far  back  afield  should  the  attempt  be 
made  to  kick  goal  to  insure  the  best  oppor- 
tunity? 

34.  The  goal  posts  are  18%   ft.  apart,  and  the  touch  down  is 
made   49   feet   from   the   nearest   post.     How   many   feet   back  is 
the  best  point  from  which  to  kick? 

35.  To  find  AB,  the  distance  across  the  river.     L  A  is  found 
to  be  107°,  ZG'  to  be  53°,  and  line  AC  to  be  24.7  rods. 


CHAPTER  IV. 

AREA  OF  POLYGONS 

353.  AREA.     Area  is   a  species  of  quantity  ( §  245 ) 
and  is  obtained  by  measuring  surface. 

354.  MEASUREMENT   OF    SURFACE.     Measurement   of 
surface  is  the  process  of  determining  the  number  of 
times  a  unit  of  surface  is  contained  in  the  given  surface. 

355.  UNIT  OF  SURFACE  OR  AREA.    A  square  having  a 
given  linear  unit  for  one  side  is  usually   taken  as  the 
unit  of  surface  or  area,  e.  g.  a  square  inch,  a  square  rod, 
etc.     There  are  exceptions  to  this,  as  the  acre  used  in 
land  measurement  which  consists  of  160  square  rods. 

356.  AREA.    The  ratio  of  a  given  surface  to  the  unit 
of  surface  is  the  area  of  the  surface.    It  is  expressed  in 
terms  of  the  unit  used,  e.  g.  15  square  inches,  24  square 
yards,  etc.    When  thus  expressed  area  tells  two  things ; 
first,  the  number  of  times  the  unit  is  contained  in  the 
surface  and  second,  the  name  of  the  unit  used. 

PROPOSITION  I. 

357.  THEOREM.     Two  rectangles  having  equal 
bases  are  proportional  to  their  altitudes. 


Given  two*  rectangles  P  and  P'  with  equal  bases  m  and 
altitudes  n  and  n'  respectively. 


170  PLANE  GEOMETRY 

To  Prove  P-=^ 
P'     ri 

The  altitudes  n  and  n'  are  assumed  to  have  a  common 

unit  of  measure.    The  case  in  which  they  do  not,  i.  e.,  in 

which  they  are  incommensurable,  will  be  considered  later. 

SUG.     1.     Suppose  n  contains  this  common  unit 

k  times  and  that  n'  contains  it  k'  times.     What 

then  is  the  ratio  of  n  to  n'  ?    Auth.  ? 

2.  Through  the  points  of  division  in  n 
and  n'  draw  lines  parallel  to  the  bases.     The  fig- 
ures thus  formed  are  rectangles.    Why  ?    They  are 
congruent.    Why  ?    Therefore  one  of  them  can  be 
used  as  a  unit  of  measure. 

3.  How  many  times   is  this  unit   rec- 
tangle contained  in  P  ?     In  P'  ?     What  is  the 
ratio  of  P  to  P'  ? 

4.  Compare  the  ratio  of  the  altitudes 
with  that  of  the  rectangle.    Auth.? 

Therefore — 

358.  COR.     //    two    rectangles   have    equal   altitudes 
they  are  proportional  to  their  bases. 

SUG.     Either  side  of  a   rectangle  may  be   re- 
garded as  the  base. 

359.  COR.     //  two  rectangles  have  a  common  dimen- 
sion, they  are  proportional  to  the  other  dimensions. 

The  dimensions  of  a  parallelogram  or  a  triangle  are  its 
base  and  altitude. 


AREA  OF  POLYGONS 


171 


PKOPOSITION  II. 

360.     THEOREM.     Tico    rectangles   are   propor- 
tional to  the  products  of  their  dimensions. 


Given  rectangles  M  and  M'  with  dimensions  a,  b  and 
a',  b'  respectively. 

M       axb 
To  Prove  —  =  • 


a'xfc' 

Proof.  SUG.  1.  Construct  a  rectangle  P  with  one 
dimension  of  M,  say  a,  and  one  dimension  of 
IT,  say  6'. 

2.  What  is  the  ratio  of  M  to  P  ?  Of  P 
to  J/'  ?    Why? 

3.  Find    the     product     of    these     two 
ratios.     What  then  is  the  ratio  of  M  to  M '  ? 

Therefore— 

PROPOSITION   III. 

361.     THEOREM.    The  area  of  a  rectangle  equal* 
the  product  of  its  base  and  its  altitude. 


Given  rectangle   P   with    dimensions   m   and   n    and 
unit  P'. 

To  Prove  area  P  =  m  x  n  times  P'. 


172  PLANE  GEOMETRY 

. 

SUG.     The   ratio  -  =  — =  HIM.     Why? 
P'     Ixl 

•'•  P  =  mnP'  or  area  P  —  mn  of  the  units  of  area. 

The  theorem  is  here  stated  in  the  usual  abbreviated  form. 
Literally  interpreted  it  would  imply  that  area  is  a  number  and 
that  the  "base  ana  altitude"  also  are  numbers.  Stated  in  full 
and  as  it  should  be  interpreted,  it  would  read:  The  area  of  a  rect- 
angle equals  the  product  of  the  measures  of  the  base  and  alti- 
tude times  the  unit  of  measure.  All  the  theorems  of  areas  will 
be  stated  in  the  abbreviated  form. 

The  name  of  a  polygon  is  frequently  used  to  designate  its 
area. 

Polygons  and  circles  are  the  surfaces  to  be  measured  in  plane 
geometry. 

362.  If  the  sides  of  the  unit  of  measure  are  exact 
divisors  of  the  corresponding  sides  of  the  rectangle  a 
simple  means  of  determining  the  area  is  to  lay  off  the 
unit  upon  the  rectangle,  noting  that  the  number  of 
times  the  base  of  the  unit  is  contained  in  the  base  of  the 
rectangle  is  the  number  of  area  units  in  one  row  and 
that  the  number  of  times  the  altitude  of  the  unit  is  con- 
tained in  the  altitude  of  the  rectangle  is  the  number  of 
rows.  Then  by  arithmetical  analysis  it  is  seen  that  the 
number  of  the  rows  times  the  number  of  the  units  in 
each  row  is  the  total  number  of  units  of  area. 

The  demonstration  of  §  357  covers  all  cases  of  practical  value 
for  if  the  altitudes  are  incommensurable  an  approximate  unit  of 
altitude  can  be  taken  as  small  as  the  needs  of  the  given  case  re- 
quire. For  instance,  if  the  altitudes  are  3  in.  and  VlTin.  the  ratio 

•> 
of   the   altitudes  will  be   approximately  —  if   1   in.   be    used   r.s 

*)/\  ^nn 

the  unit;  —  if   the  unit   is   .1    in.;  —  if  the  unit   i»  .01    in.; 

etc.  The  answer  in  this  last  case  is  correct  to  less  than  the  area 
of  a  strip  the  length  of  the  rectangle  and  .01  of  the  linear  unit 
in  width.  Thus  the  accuracy  of  the  area  computed  will  depend 


AREA    OF    POLYGONS  173 

upon  the  accuracy  of  the  measurement  of  the  dimensions,  for  the 
demonstration  by  §  357  can  be  carried  rigorously  to  an  approxi- 
mation far  beyond  the  skill  of  man  to  measure  in  any  given  case. 
All  actual  measurement  gives  but  an  approximation  of  the  true 
value  of  the  magnitude  and  the  closeness  of  the  approximation 
in  measurement  is  usually  determined  by  the  degree  of  accuracy 
required. 

PROPOSITION  IV. 

363.     THEOREM.     The  area  of  a  parallelogram 
equals  the  product  of  its  base  and  altitude. 


Given  Parallelogram  DEFG  with  base  b  and  alti- 
tude a. 

To  Prove  area  of  DF  =  ab. 
Proof.     SUG.  1.     From   two   adjacent   vertices   as   D 

and  E,  drop  perpendiculars  to  the  opposite  side, 

as  DM  and  EN. 

2.  Compare  A  DMG  with  A  EXF. 

3.  Compare  the  areas  of  the  parallelo- 
gram DF  and  the  rectangle  DN. 

4.  What  is  the  area  of  DF<!     QiDNl 
Therefore — 

Does  the  above  demonstration  apply  to  both  figures? 

Construct  carefully  a  parallelogram,  ABCD,  with  sides  from 
three  to  five  inches  in  length.  Construct  the  altitude  upon  AB 
as  a  base  and  compute  the  area.  Construct  the  altitude  upon  AD 
as  base  and  compute  the  area.  Compare  the  results  as  a  test  for 
both  accuracy  of  construction  and  accuracy  01  measurement. 


174  PLANE  GEOMETRY 

PROPOSITION  V. 

364.     THEOREM.     The  area  of  a  triangle  equal* 
one-half  the  product  of  its  base  and  altitude. 


Given  a  A  EFG,  with  base  b  and  altitude  a-. 

To  Prove  area  of  EFG  =  —  • 

2 

Proof.  SUG.  1.  Through  vertex  F  draw  a  line  par- 
allel to  base  EG,  and  equal  to  EG.  Connect  its 
extremity  H  with  G.  What  kind  of  a  figure,  EH, 
is  thus  formed?  Why? 

2.  What  is  the  area  of  EH  ?     Compare 
the  given  triangle  with  EH.     What  is  the  area 
of  the  triangle? 

3.  Compare  the  bases  and  the  altitudes 
of  the  triangle  and  EH. 

Therefore— 

365.     COR.   I.     Two   triangles  having   the   same   alti- 
tude are  proportional  to  their  bases. 

AP      axb       b 
SUG.  —  —  —  Auth. 

AP'     axb'      b' 


366.  COR.  II.  Two  triangles  having  equal  bases  are 
proportional  to  their  altitudes. 

Proof  left  to  the  pupil. 

Draw  the  three  altitudes  of  a  triangle  and  from  each  make  the 
necessary  construction  for  the  proof  of  Prop.  V. 

Construct  a  triangle  and  its  three  altitudes.  Make  the  meas- 
urements and  compute  the  area  from  each  altitude.  Compare 
the  results  as  a  check. 


AREA    OF    POLYGONS  175 

PROPOSITION    VI. 

367.  THEOREM.    Any  two  triangles  are  propor- 
tional to  the  products  of  their  two  dimensions. 

Given  A  P  and  P'  with  dimensions  a,  b  and  a',  b' 
respectively. 

P        ab 
To  Prove  r  =  — ' 

P       ab 

Proof.     SUG.     Find  the  ratio  of  the  areas  and  sim- 
plify the  expression. 

PROPOSITION  VII. 

368.  THEOREM.    Two  triangles  having  an  angle 
in  one  equal  to  an  angle  in  the  other  are  propor- 
tional to  the  products  of  the  sides  including  the 
equal  angles. 


^G 

Given  A  EFG  and  A  E'F'G'  with  LE=LW. 

To  Prove  — — -  —  =  - 

A  E'F'G'     E'F'xE'G' 

Proof    SUG.  1.  Place  A  EFG  upon  A  E'F'G'  so    that 
EF    falls    on  E'F',    EG  on  E'G',  and#  on  E'. 
2.     Note   that    A    E'F'G'    and    E'F'G 
have  a  common  altitude,  whence 
AE'F'G  &E'FG 


-  —  ?    Similarly 
'' 


kE'F'G  AE'F'G 

3       kE'FG  =  ? 

^  E'F'G'" 
Therefore — 


17G  PLANK  GEOMETRY 

1.  Compare  the  areas  of  three  triangles  formed  by  drawing 
lines  through  the  vertex  of  a  given  triangle  so  as  to  trisect  the  base. 

2.  If   the   base   of  a   triangle   remain   unchanged  while   the 
vertex  is  moved  in  a  line  parallel  to  the  base,  what  is  the  effect 
upon  the  area  of  the  triangle? 

3.  Divide   a  triangle   into   two   parts  by   a  line   through  the 
vertex  so  that  one  part  is  three  fifths  of  the  other. 

4.  Through  a  given  point  in  one  side  of  a  triangle  draw  a 
line  bisecting  the  triangle. 

5.  Through  a  given  point  in  one  side  of  a  triangle  draw  a 
line  which  will  cut  off  one  third  of  the  triangle. 

6.  Divide  a  triangle  into  two  equal  parts  by  a  line  parallel 
to  the  base. 

7.  Divide  a  triangle  into  two   parts  by  a  line    parallel  to  the 
base  so  that  one  part  is  one  third  the  other. 

8.  If  from  any  point  in  a  parallelogram  lines  be  draAvn  to  the 
four  vertices  the  sum  of  either  pair  of  the  opposite  triangles  thus 
formed  equals  one  half  the  parallelogram. 

9.  The  area  of  a  square  or  a  rhombus  equals  one  half  the 
product  of  the  diagonals. 

10.  If  EF  and  DG  are  diagonals 
of    a    parallelogram    prove    that    any 
pair  of  triangles,  as  P  and  P',  on  the 
same  side  of  a  diagonal  are  equal. 

11.  Let  M   (or  M')     be  any  point      on    diagonal  EF   of   the 
preceding  figure    (or   on   EF  produced).      Connect  M  with  E  and 
F.     Prove  A  MEO  =  A  MFO-  and  A  M'EO=&  M'FO. 

12.  Drive    three    stakes   several   rods    apart   so    as    to    form    M 
triangle.        Measure    the    three    sides    and    compute     the    area. 
Measure   an   altitude  and  base  and   compute  the   area.     Compare 
the    results.       If    the    several     rrsuhs     dilTer    to     a    considerable 
amount    repeat    the    work    to    find    the    error. 

13.  Draw    to     some    convenient    scale    the     triangle     of    the 
preceding  exercise.     Compute   the   area  by  measurements  accord- 
ing to  one  or  other  of  the  methods  used  before  and  compare  the 
result  with  those  derived  from  the  direct  measurements. 


AREA    OF   POLYGONS  177 

PROPOSITION    VIII. 

369.     PROBLEM.    To  determine  the  area  of  a  tri- 
angle in  terms  of  its  sides. 


Given  A  ABC  with  sides  a,  ~b,  c  and  altitude  fta  on 
side  a. 

To  find  area  of  ABC  in  terms  of  a,  ~b,  c. 

SUG.     AABC=   -^       In  this  substitute  for 

a 

h&  its  value  in  terms  of  a,  b,  c,  and  simplify.  §340 

1 .  Write  the  formula  for  Prop.  V  using  d  and  then  c  as  base. 
Show  that  the  simplified  formulas  are  the  same  irrespective  of  the 
side  used  as  base. 

2.  The  sides  of  a  triangular  piece  of  ground  are  7,  9,  and  12 
rods.     Find  the  area. 

3.  Construct    a    triangle   having    sides    of    8,    10,    12    inches. 
Find  its  area  by  §  369.     Measure   one  altitude  and  compute  the 
area.     Check  results  by  comparison. 

4.  The    area    of    a    triangle    circumscribed    about    a    circle 
equals   one  half  the  product   of  the   perimeter  and   the  radius   of 
the    circle.      Prove    the    same    theorem    also    for    a    circumscribed 
polygon. 

5.  A  number  of  triangles  have  as  bases  equal  sects  of  the 
same  straight  line.  If  the  opposite  vertices  are  at  the  same  point, 
which  triangle  is  the  larger? 

6.  Several   triangles   have   as    bases    equal    sects    of   a    given 
straight  line.     If  their  areas  are  to  be  equal  what  is  the  locus  of 
the  opposite  vertices? 

7.  Connect  the  mid-points  of  the  adjacent  sides  of  a  paral- 
lelogram.    Prove    (1)    that  the  included  quadrilateral  is  a  paral- 
lelogram,   (2)    that   it    equals   one   half   the    given   parallelogram, 
(3)   that  the  triangles  at  the  four  vertices  are  equal. 


178 


PLANE  GEOMETRY 


PROPOSITION    IX. 

370.     PROBLEM.     To  construct  a  triangle  equal 
to  a  given  polygon. 


Given  any  polygon  as  DEFG 

To  Construct  a  triangle  equal  to  DEFG . .  . 

Construction.  SUG.  1.  Draw  a  diagonal  cutting  off 
a  triangle  as  EFG  and  through  the  vertex  F 
draw  a  line  parallel  to  the  diagonal  meeting  DE 
produced  at  M .  Join  G  and  M. 

2.  Prove    A  EMG  =  A  EFG. 

3.  Prove  DEFG .  . .  =  DEMG, , . 

4.  Compare  the  number  of  sides 
in  DEFG..  .  and  DEMG... 

5.  Proceed  with  the  new  poly- 
gon DEMG.  . .  as  with  DEFG. . . 

6.  A  triangle  is  finally  obtained 
by  this  process.     Why  does  it   equal  the   ghen 
polygon  ? 

PROPOSITION    X. 

371.  THEOREM.  The  area  of  a  trapezoid  equal* 
one-half  the  product  of  the  altitude  and  the  sum 
of  the  bases. 


Given  the  trapezoid  DEFG  with  altitude  h  and  bases 
DE  and  FG. 


A    OF    POLYGONS  179 


To  Prove    Area    DEFG=         (DE  +  FG)  h. 

Proof.     Suo.  1.       Area  of   &DEF=--DE.     What 

2 

is  the  area  of  A  FGD  ? 

2.     What  is  the  area  of  DEFG  ? 
Therefore— 

What  is  the  unit  of  addition  in  step  2  ?  What  are  the  co- 
efficients? Verify  the  addition  algebraicly. 

DEx^+FGx  ~-.=  (DE  +  FG)x  £-. 

372.  AREA  OF  A  POLYGON.  Various  methods  are  used 
in  finding  the  areas  of  irregular  polygons,  among  which 
the  following  may  well  be  noticed : 

From  any  vertex  of  the  polygon  draw  all  possible 
diagonals.  The  polygon  is  by  this  means 
divided  into  triangles.  The  sides  and  alti- 
tudes of  these  triangles  may  be  measured 
and  their  areas  computed  by  any  of  the 
methods  already  shown.  The  area  of  the  entire  polygon 
is  then  found  by  addition. 

Another  method  is  to  draw  a  diagonal,  preferably 
the  longest  one  and  from  the  remaining  vertices  drop 
perpendiculars  upon  this  diagonal.  The  ^^\  A 
polygon  is  thus  divided  into  triangles  and  (/"" ^t"-T-L.\ 
trapezoids.  If  the  bases  and  altitudes  of 
these  triangles  and  trapezoids  are  measured 
their  areas  can  be  computed  and  the  area  of  the  polygon 
thus  obtained. 

Another  method  is  to  draw  through  any  vertex  of  the 
polygon  a  straight  line  exterior  to  the  polygon  upon 


180 


PLANE  GEOMETRY 


which  perpendiculars  are  dropped  from  the  remaining 
vertices.     In  this  way  triangles  and  trape- 
zoids  are  formed.     The  bases  and  altitudes 
of  these  are  measured  and  their  areas  com- 
puted.    The  ar^a  of  the  given  polygon  is  "^-^ } 

found  by  subtracting  the  areas  of  the  parts  exterior  to 
the  polygon  from  the  sum  of  the  areas  of  all  the  parts. 
Solve  Ex.  1  P.  184. 

PROPOSITION  XL 

373.  THEOREM.  The  ratio  of  the  areas  of  tivo 
similar  triangles  is  equal  to  the  ratio  of  the 
squares  of  any  two  homologous  sides  or  homolo- 
gous altitudes. 


M  M' 

Given  two  similar  A,  ABC  and  A'B'C'  with  homol- 


ogous altitudes  AM  and  A'M' . 
To  Prove 


«      * 
Proof 


AB 


C'A'2     AB'*     AfM'' 

BCxAM 

Tfr  hy  ? 


BC       AM      /f 

= X  -  (factoring). 

B'C'     A'M' 


B^=CA^=AB^=AM_ 
"  B'C'     C'A'      A'B'      A'M' 


ABC        B 


* 


~AM 


3.    .-.  ---  =  ==-•  =  =-  etc.  \\liy? 
A'B'C'     B'C'*      A'M'2 


AREA    OF   POLYGONS  181 

PROPOSITION  XII. 

374.  THEOREM.  The  ratio  of  the  areas  of  iwo 
similar  polygons  is  equal  to  the  ratio  of  the 
squares  of  any  two  homologous  sides. 

Given  two  similar  polygons  P  and  P'  with  AB  and 
A 'B'  any  pair  of  homologous  sides. 

P' 
To  Prove          = 


Proof.  SUG.  1.  From  vertices  A  and  A'  draw  all 
possible  diagonals  dividing  P  and  P'  each  into 
the  same  number  of  triangles,  similar  each  to 
each  and  similarly  placed. §  286  (1), 

2.  The  ratio  of  similitude  of  each  pair 
of  similar  triangles  is  the  same  as  the  ratio  of 
similitude  of  P  and  P' .    For : 

3.  Let  A1?  A/;  A2,  A2';  A,,  A,';  etc., 
represent  the  pairs  of  similar  triangles  into  which  P  and 
P'  are  respectively  divided.     Then 
AL_A_a__A_a_  _^!_.    \^h    9 
A/~A2'~A3'~  ~A^'2' 

4.  Then 

P       A.  +  A,  +  A,  +  .  Z1J2 

—  =  — ' —  — — 'See  §317. 

T*'  A        '      I          A        '    I          A        '      I  Af  D'^ 

Therefore— 

375.  State  each  of  the  cases  in  which  the  ratio  of 
areas  has  been  found.  Notice  that  the  ratio  of  areas  is 
always  expressed  by  the  product  of  two  factors.  If 
there  is  a  common  dimension,  this  factor  can  be  can- 
celled out  of  both  terms  of  the  ratio.  State  all  the  cases 
thus  far  demonstrated  in  which  this  is  true;  also  those 


182 


PLANE  GEOMETRY 


in  which  the  ratio  remains  a  product.    Under  what  con- 
ditions may  this  product  be  written  as  a  square? 

PROPOSITION  XIII. 

376.  THEOREM.  The  square  described  upon  the 
hypotenuse  of  a  right  triangle  is  equal  to  the  sum 
of  the  squares  described  upon  the  two  other  sides. 


Given  A  ABC  with  L  B  a  rt.  angle  and  AE,  CF,  and 
BH  squares  upon  the  sides  AC,  CB,  and  BA  respec- 
tively. 

To  Prove   AE  =  CF  +  BH. 

Proof.  SUG.  1.  Draw  BG  II  to  CE,  meeting  AC  in 
G  and  NE  in  0.  Draw  BN,  BE,  CH  and  AM. 
What  kind  of  polygons  are  CEOG  and  AGON*! 
Why? 

2.  Compare  A  BCE  and  ACM. 

3.  ABF  is  a  straight  sect.     Why  ? 

4.  Compare  the  area  of  A  ACM  with 
the  area  of  the  square  BM;  the  area  of  A  BCE 
with  the  area  of  the  rectangle  CO. 

5.  Compare  the  area  of  the  square  BM 
with  the  area  of  the  rectangle  CO. 

6.  Compare  A  CAH  and  NAB. 

1.     CBD  is  a  straight  sect.    Why? 
8.     Compare  the   square   BH  with   the 
rectangle  AO. 


AREA    OF    POLYGONS  183 

• 
« 
9.     Compare  the   sum   of  the  areas   of 

the  squares  BM  and  BH  with  the  sum  of  the  areas 
of  the  rectangles  CO  and  AO,  i.  e,  with  the  area 
of  the  square  CN. 
Therefore— 

Compare  this  theorem  with  §  330  and  show  that  the 
theorems  are  the  same  except  that  one  is  demonstrated 
by  algebraic  processes  and  the  other  by  geometric  proc- 
esses. One  deals  purely  with  the  number  and  the  other 
with  quantity. 

377.  COR.  The  square  upon  the  leg  of  a  right  tri- 
angle equals  the  square  upon  the  hypotenuse  minus  the 
square  upon  the  other  leg. 


378.  SCHOLIUM.  This  proposition  is  known  as  the 
Pythagorean  proposition.  It  is  named  in  honor  of 
Pythagoras,  570  B.  C.,  who  is  supposed  to  have  given 
the  first  demonstration  of  it.  The  above  demonstration 
is  usually  associated  with  his  name. 

The  following  is  one  among  the  many  interesting 
proofs  which  have  been  devised  for  this  proposition: 

Let  ABC  be  the  right  A,  with  AG  the 
square  upon  the  hypotenuse.     Draw  EF 
||  AB,  GF  ||  CB,  and  EH  l  AB. 
Why? 
=  sq.  EC.    Why  ? 

Also  A  EAH  =  A  ABC.    Why  ? 

•'•  EH  =  AB  and  area  of  O  EB  =  area  of  sq.  on  AB. 
Why? 

In  the  same  way  O  BG  =  sq.  on  BC. 


184 


PLANE  GEOMETRY 


1.  Construct    with    instruments    a    pentagon.     Compute    the 
area  by  §§  370  and  372.     The  comparison  is  a  test  of  construction. 

2.  Find  the  area  of  a  triangle  with  sides  of  8,  12,    15. 

3.  If  the  diagonals  of  a  quadrilateral  intersect  at  right  an- 
gles, show  that  the  area  of  the  quadrilateral  is  one  half  the  area 
of  the  rectangle  the  sides  of  which  are  equal  to  the  diagonals  of 
the  quadrilateral. 

4.  If  three  equal  circles  are  tangent  each  to  the  two  others, 
the  lines  joining  their  centers  form  an  equilateral  triangle. 

5.  Construct  a  triangle  with  an  area  9  times  that  of  a  tri- 
angle the  sides  of  which  are  6,  7,  9. 

6.  A  puzzle.     Draw  on  a  larger  scale  the 
accompanying  figure.    Cut  it  into  5  pieces  along 
the  dotted  lines  and  rearrange  the  pieces  form- 
ing a  square  on  the  line  M N.     This  is  an  illus- 
tration of  what  theorem? 

7.  A  puzzle.     Construct  a  square  MN  of 
any  size.     Draw  MD  at  random.     Drop  perpen- 
diculars EF  and  GH  to  MD.     Make  HP  equal 
to  ND  and  drop  a  perpendicular  from  P  to  MD. 
The  square  MN  can  now  be  cut  into   o  parts 
which    can    be    rearranged    into    two    squares. 
This  illustrates  §  3-76. 

8.  Verify    the    theorem    of    §   376   by   making    the 
upon  paper  ruled  in  squares. 

Sue.  Construct  the  squares  on  the  two  legs 
and  on  the  hypotenuse  of  the  triangle  and  count 
the  small  squares  included  in  each. 

9.  In  the  accompanying  figure  AM  is  a  rt. 
A,  and  A  N,  D,  and  P  are  equilateral.     Prove 


N 


M 


}      5  —  ^1  Take  proportions  2  and  3  by  alternation  and 


compare,  hence 


D  +  p      d*+p* 
-  —  • 


EXERCISES  185 


Also  ^  =  '^  and  hence  bv  division  —  ±_  -  —  ^-  =1. 
.-.     A  Z>  +  A  P  =  A  tf  . 

Is  this  relation  still  true  if  for  N,  D,  P  we  substitute  any  simi- 
lar polygons? 

10.  The  square  on  the  side  of  a  triangle  opposite  an  obtuse 
angle  equals  the  sum  of  the  squares  upon  the  two  other  sides  plus 
twice  the  rectangle  formed  by  one  side  and  the  projection  of  the 
other  side  upon  that  side. 

Givfix  A  ^BC  with  /.B  obtuse,  AE  the  sq.  opposite  L  B,  with 
BM  and  BE  the  squares  on  the  two  other  sides,  BB'  the  projection 
of  AB  upon  BC,  and  B'F  the  rectangle. 
To  PROVE  AE  =  BM  -f  BH  +  2  B'  F. 

N 


D'          3T?       M 

Sug.  1.     Assume  the  theorem  in  its  algebraic  form  from 
§  338.  Substitute  for  the  algebraic  products  the  areas  which 
they  represent  in  the  above  figure.      (Draw    BE,     CD    and 
extend  AB  to  D'C.) 

SUG.  2.  From  each  vertex  of  the  given  triangle  drop 
perpendiculars  to  the  farther  side  of  the  opposite  squares, 
or  to  these  sides  extended  if  need  be.  Connect  the  vertices 
of  the  triangle  with  the  opposite  vertices  of  their  respective 
opposite  squares.  Work  out  a  demonstration  similar  to 
§376. 

]  1 .     The  square  upon  the   side  opposite  an  acute  angle  of  a 
triangle  equals  the  sum  of  the  squares  upon  the  other  two  sides 
minus  twice  the  rectangle  formed  by  one  of  those  sides  and  the 
projection  of  the  other  upon  it. 
GIVEN  A  ABC  with  sides  a,  b,  c. 
To  PROVE  the  relation  O2z=&2-f-c2 —  2bm,  in  which  0,2,  faj  c* 


186 


PLANE  GEOMETRY 


represent  squares  on  the  sides   a,  fe,  c,  respectively  and   fern  the 
rectangle  formed  by  side  fe  and  the  projection  of  c  on  &. 

SUG.  1.     Prove  from  §  376  as  in  preceding  exercise. 

Sue.  2.     Draw  the   auxiliary  lines   and  prove  geometric- 
ally   as  in  the  preceding  exercise. 

12.  What  is  the  ratio  of  &MNO  to  A  PMO,  ZM  being  90°? 
Fig.  1. 

13.  Prove  geometrically   (a  +  6)2  =  a2  +  2afe  +  62.     Fig.  2. 

14.  Prove  geometrically  (a — fe)2  =  a2 —  2afe  +  &2.     Fig.  3. 

15.  Prove  geometrically   (a  +  fe)    (a  —  fe)  —  a-  —  fe2.     Fig.  4. 


a       b 

fl-6 

b 

... 

r/          " 

"'  » 

i 

b 

*! 

1  f  -b 

Fig.  1.          Fig.  2.  Fig.  3.  Fig.  4. 

16.  Prove  by  a  figure  the  following  formulas: 

a  (a  —  fe)  —  «2  — -ab 

a  (fe  +  c)  =  afe  +  ac 

(a  +  fe)    (a  +  c)  =  a2  +  a&  +  ac  -f  fec. 

(ct -f- &)  (a  —  c)izr«2-|-a6 — ac — fee 

(a — &)   (a  —  c)  =  a2  —  a&  —  ac  +  fee. 

(a-f  fe)3  -f  (a  —  6)2  =  2:ia  +  2fe2 

(a  +  6)    (c  +  d)  =  ac  +  ad  +  fee  +  fed. 

1 7.  To    construct    a    square    equal    to    the   sum    of    two    given 
squares. 

GIVEN  a  square  on  a  and  a  square  on  fe.    Fig.  5. 
To  CONSTRUCT  a  square  equal  to  #2  -f  fe2. 

SUG.     Construct  a  right  triangle  of  which  a  and  fe   are 

the  legs.     Complete  construction.     The  proof  is  left  to  the 

pupil. 

18.  Construct    a    square    equal    to    the    sum    of    three    given 
squares;  of  four  given  squares;  of  n  given  squares. 

19.  Construct   a   square   equal  to    the    difference   between   two 
given  squares. 


PKOBLEMS    OF    CONSTRUCTION 


187 


PROPOSITION  XIV. 

379.     PROBLEM.      Upon  a  given   base   to   construct  a 
rectangle  equal  to  a  given  rectangle. 


Given  rectangle  DF  with  adjacent  sides  b  and  c  and 
sect  a. 

To  Construct  a  rectangle  on  a  equal  to  DF. 

SUG.  1.     Let  x  represent  the   altitude   of  the 
required  rectangle.     Then  b  X  c  =  a  X  x. 

2.  From  this  derive  a  proportion  with  x 
as  the  fourth  and  unknown  term. 

3.  Find  x  and  complete  the  construction. 

4.  Show  that  the   constructed  rectangle 
equals  DF. 

PROPOSITION  XV. 

380.     PROBLEM.     Upon  a  given  sect  as  base,  to  con- 
struct a  rectangle  equal  to  a  given  square. 


Given  sect  b  and  a  square  on  sect  a. 
To  Construct   on   b    a   rectangle   equal  to  the   given 
square. 

SUG.  1.     Let  x  be  the  altitude  of  the  required 
rectangle.    Then  a2  =  b  X  x. 

2.  From  this  derive  a  proportion  with  x 
as  the  fourth  and  unknown  term.  What  prob- 
lem is  involved  in  the  finding  of  xf 

3.  Complete  the  construction  and  prove 
the  constructed  rectangle  equal  to  the  given 
square. 


188  PLANE  GEOMETRY 

1.  Upon  a  given  sect  as  base  construct  a  rectangle  the  area 
of  which  shall  equal  the  sum  of  the  areas  of  a  given  square,  a 
given  trapezoid,  and  a  given  triangle. 

2.  The  area  of  a  square  inscribed  in  a  circle  is  one-half  the 
area  of  any  square  circumscribed  about  the  circle. 

3.  Construct  a  parallelogram  with  a  given  base  and  a  given 
angle,  the  area  of  which  shall  equal  that  of  a  given  rectangle. 

4.  Find  the  dimensions  of  a  rectangle  the  perimeter  of  which 
is  26  in.  with  an  area  of  40  sq.  in. 

PROPOSITION  XVI. 

381.     PROBLEM.    To  construct  a  square  equal  to 
a  given  rectangle.    . 

-I    •>     I 

Given  a  rectangle  with  adjacent  sides  a  and  b. 
To  Construct  a  square  equal  to  rectangle  ab. 

SUG.  1.     Let  x  represent  the  side  of  the  re- 
quired square.     Then  x2  =  ab. 

2.  Derive  a  proportion  and  find  x. 

3.  Complete  the  construction. 

5.  Verify  the  constructions  in  §§  379,  380,  and  381  by  meas- 
uring the  dimensions  and  computing  the  areas. 

6.  Construct  a  square  equal  to  a  given  trapezoid. 

7.  Construct  a  square  equal  to  a  given  triangle. 

8.  Construct  a  square  equal  to  a  given  parallelogram. 

9.  If  the  hypotenuse   of  a  right  triangle  is  15   ft.  and  the 
ratio  of  its  legs  is    £     what  is  its  area? 

10.  Cut   off   one-third  of  a  parallelogram   by   a   line   through 
a  vertex.     Cut  off  one-fourth  in  the  same  manner. 

11.  Construct  an  isosceles  triangle  having  a  given  altitude  and 
equal  to  a  given  triangle. 

12.  Bisect  a  parallelogram  by  a  line  parallel  to  a  given  line. 

13.  Construct  a  triangle  equal  to  a  given  triangle,  M,  on  base, 
o,  and  with  its  opposite  vertex  in  a  given  line,  x.    Can  this  line  be 
parallel  to  the  base? 


EXERCISES  189 

14.  The  diagonals  of  two  squares  are  5  ft.  and  9  ft.  respect- 
ively.    What  is  the  diagonal  of  a  square  equal  to  their  sum? 

15.  Divide  a  triangle  by  a  line  through  trisection  points  on 
one  side  into  parts  proportional  to  1,  2,  3. 

16.  The   side  of  an  equilateral  triangle  is   10.     What  is  tbi 
length  of  a  side  of  a  regular  hexagon  of  the  same  area?     Of  a 
square  of  the  same  area? 

17.  The   diagonal  of  a  rectangle   is  10  and  the  ratio  of  the 
sides  is  f.     Find  the  area.     If  the  diagonal  is  40  and  the  ratio 
of  the  sides  is  J.,  find  the  area. 

18.  The  difference  between  the  squares  of  two  sides  of  a  tri- 
angle equals  the  difference  between  the  squares  of  the  projections 
of  those  sides  upon  the  third  side. 

19.  If  two  circles  are  tangent,  internally  or  externally,  chords 
of  the  one  passing  through  the  point  of  contact  are  divided  pro- 
portionally by  the  other. 

20.  Draw  a  chord  through  a  given  point  within  a  circle  which 
shall  be   divided  by   this  point  in  the   ratio    1         Also   the  ratio 
TO 

Sue.    Use  preceding  ex. 

21.  Draw  a  secant  from  a  point  without  a  circle  terminated 
by  the  circle  such  that  its  ratio  to  its  external  segment  s.hall  be 

3  to  1.    Also    — .       Is  this  problem  ever  impossible? 

22.  Through  a  given  point  draw  a  line  so  that  its  distances 
from    two  given  points  shall  be  in  a  given  ratio. 

23.  What  is  the  locus  of  points  which  divide  all  chords  of  a 
circle  passing  through  a  given  point,  internally  or  externally,  in 
a  given  ratio? 

24.  Let  DEFG ...  be  any  polygon.         Jo4  n  the 
vertices  to  a  point  O.     Through  D't   any  point  on 
OD,   draw  a   line  parallel   to   DE     terminating  in 

E'  on  DE.  Through  E'  draw  a  line  parallel  to 
EF,  etc.  Prove  the  polygon  D'  E'  F'  G'  ...  similar 
to  DEFG. .  . 


190  PLANE  GEOMETRY 

Two  similar  polygons  can  always  be  placed  in  such  a  position 
with  respect  to  some  fixed  point. 

This  property  is  characteristic  and  is  sometimes  taken  as  the 
definition  of  similarity^  The  point  in  which  the  lines  joining 
homologous  vertices  meet  is  called.'  the  center  of  similitude. 

24.  The  area  jf  a  triangle  equals  one  half  the  product  of  two 
sides  and  the  sine  of  the  included  angle. 

This  proof  covers  only  the  case  in  which  the  included  angle 
is  acute.  The  general  theorem  is  proved  in  trigonometry. 

GIVEN  A  ABC  with  sides  a  and  c  including  the  acute  angle  B. 

To  PROVE  area  ABC  =  %  ac  sin  B. 

PROOF.  Area  ABC  -  \  ali&  and  h&  =  c  sin  B.  Hence  the 
theorem.  Fig.  §  369. 

25.  Drive  three  stakes  in  the  School  grounds  so  as  to  form  an 
acute    triangle.     Measure  two  sides  and  the   included   angle  and 
compute  the  area  by  ex.  24,  also  measure  a  base   and    altitude 
and  compute  and  compare  results,  Review  your  work  if  not  approx 
imately  the  same. 


CHAPTER  V. 

REGULAR  POLYGONS. 

382.  REGULAR  POLYGON.     A  polygon  which  is  both 

equilateral  and  equiangular  is  regular. 

The  equilateral  triangle  and  the  square  are  regular  polygons. 

PROPOSITION  I. 

383.  THEOREM.     An    equilateral    polygon    in- 
scribed in  a  circle  is  a  regular  polygon. 


Given  AD,  an  equilateral  polygon  inscribed  in  a  circle. 

To  Prove  that  AD  is  a  regular  polygon. 

Proof.     Sue.  1.     According  to  the  definition,  §  382, 

what  remains  to  be  proved  in  order  that  AD  be 

regular  ? 

2.  By  what  may  the  angles  A,  B,  C, 
etc.,  be  measured? 

3.  How    do   they   compare   with   each 
other? 

4.  Test    AD    by    the    definition    of   a 
regular  polygon. 

Therefore — 

384.  COR.  //  a  circle  be  divided  into  any  number  of 
equal  parts,  the  lines  joining  the  points  of  division  form 
a  regular  polygon. 

Proof  left  to  the  pupil. 


192  PLANE  GEOMETRY 

PROPOSITION  II. 

385.  THEOREM.  A  circle  can  be  circumscribed 
about  a  regular  polygon.  A  circle  can  be  inscribed 
in  a  regular  polygon. 


Given  a  regular  polygon,  AD. 

To  Prove.    I.     A  circle  can  be  circumscribed  about 
AD. 

SUG.  1.  At  M  and  N  the  mid  points  of  two 
adjacent  sides  erect  perpendiculars  extended  un- 
til they  meet  as  at  0.  Why  do  they  meet? 

2.  Join  0  to  the  vertices  A,  B,  C,  etc. 
Compare  4   AOE  and  BOC ;  L  1,  2  and  3. 

3.  A    DOC   =   A    COB.     Why?     Hence 
OD  =  OB.    Why  ? 

4.  Similarly  all  the  sects  OA,  OB,  OC, 
OD,  etc.,  are  equal  and  hence  a  circle  with  center 
0  can  be  circumscribed.    Why? 

To  Prove.     II.     SUG.  1.     What  was  proved  concern- 
ing the  successive  A  AOB,  BOC,  etc.? 

2.  Compare     the     sects     OM, 

ON,  etc.     Auth. 

3.  Complete     the     demonstra- 
tion. 

Therefore— 

Query.    How  many  sides  in  the  polygon  .1  />  ?    See  the 
theorem. 


REGULAR  POLYGONS  193 

386.  RADIUS  OF  A  REGULAR  POLYGON.    The  radius  of 
the  circumscribed  circle  is  the  radius  of  a  regular  poly- 
gon. 

387.  APOTHEM  OF  A  REGULAR  POLYGON.     The  radius 
of  the  inscribed  circle  is  the  apothem  of  a  regular  poly- 
gon. 

388.  CENTER  OF  A  REGULAR  POLYGON.    The  center  of 
the  inscribed  and  circumscribed  circles  is  the  center  of 
a  regular  polygon. 

389.  ANGLE  AT  THE  CENTER  OF  A  REGULAR  POLYGON. 
The  angle  formed  by  two  radii  drawn  to  two  adjacent 
vertices  of  the  polygon  is  the  angle  at  the  center  of  a 
regular  polygon. 

From  the  definitions  just  given  the  following  corol- 
laries can  be  deduced.    The  pupil  should  prove  them. 

390.  COR.  1.     The  angle  at  the  center  of  a  regular 
polygon  is  equal  to  four  right  angles   divided  by   the 
number  of  sides  of  the  polygon. 

391.  COR.  II.     An  interior  angle  of  a  regular  poly- 
gon is  equal  to  the  sum  of  all  the  interior  angles  of  the 
polygon  divided  by  the  number  of  sides  of  the  polygon. 

392.  COR.  III.     The  angle  at  the  center  of  a  regular 
polygon  equals  an  exterior  angle  and  is  the  supplement 
of  an  interior  angle  of  the  polygon. 

393.  COR.  IV.     The  radius  of  a  regular  polygon  bi- 
sects the  angle  of  the  polygon  to  which  it  is  drawn. 

1.  A  farm  is  320  rods  long,  and  rectangular  in  shape.     From 
one  end  a  square  farm  is  cut  off,  leaving  120  acres.     How  wide  is 
the  farm  and  how  many  acres  in  the  entire  piece? 

2.  From  any  point  M  in  side  BC  of  A  ABC  draw  c,  sect  meet- 
ing AB  produced  in  D,  so  that  MD  is  bisected  by  AC.     Also,  so 
that  MD  is  divided  by  AC  in  any  given  ratio. 


194  PLANE  GEOMETRY 

PROPOSITION   III. 

394.  THEOREM.  //  a  circle  is  divided  into  any 
number  of  equal  parts,  the  tangents  drawn 
through  the  points  of  division  form  a  regular  cir- 
cumscribed polygon. 


Given  the  circle  ABC .  . .  . ,  divided  into  equal  parts 
AB,  CD,  EF. .  .,  with  tangents  at  the  points  A,  B,  C,.  .  . 
forming  the  polygon  A  'B  '€' .  .  .  . 
To  Prove  that  A'B'C' ...  is  regular. 

SUG.  1.     A  A'  AB,  B'BC,  etc.,  are  isosceles  and 
congruent.     Why  ? 

2.  Compare  AA' ,  A'B,  BB' ,  B'C,  etc. 

3.  Prove  A'B'  =  B'C'  =  C'D',  etc. 

4.  Prove  A.  A' ,  B' ,  C',  etc.,  equal. 

5.  Apply    the    definition    of    a    regular 
polygon. 

Therefore— 

395.  COB.  I.     //  the  vertices  of  a  regular  inscribed 
polygon  are  connected  with  the  mid  points 

of  the  arcs  subtended  by  the  sides,  a  regu- 
lar inscribed  polygon  of  double  the  number 
of  sides  is  formed.  §  394. 

396.  COB.  II.     The  perimeter  and  the  area  of  a  regu- 


REGULAR  POLYGONS  195 

lar  inscribed  polygon  are  less  than  the  perimeter  and 
area  respectively  of  a  regular  inscribed  polygon  of  dou- 
ble the  number  of  sides. 

397.  COR.  III.     //    a    regular    polygon    is    circum- 
scribed about  a  circle  and  tangents  are  drawn  at  the 
mid  points  of  the  intercepted  arcs,  a  regular  circum- 
scribed polygon  of  double  the  number  of  sides  is  formed. 
§  394. 

398.  COR.    IV.     The  perimeter  and   the   area   of  a 
regular  circumscribed  polygon  are  greater  than  the  peri- 
meter and  the  area  respectively  of  a  regular  circum- 
scribed polygon  of  double  the  number  of  sides. 

1.  The  ratio  of  similitude  of  two  similar  polygons  is    A    and 
the  sum  of  their  areas  is  518  sq.  in.    Find  the  area  of  each. 

2.  Find  the  base  of  a  rectangle  with  an  area  of  108  sq.  ft. 
and  an  altitude  of  6  ft.     Compare  the  method  with  that  of  379. 

3.  Find  the  area  of  a  right  triangle  with  an  hypotenuse  of 
1  ft.  8  in.  and  one  leg  of  1  ft.  in  length. 

4.  The  area  of  a  circumscribed  polygon  equals  one-half  the 
product  of  its  perimeter  and  the  radius  of  the  circle. 

5.  If  the  mid  points  of  two  adjacent  sides  of  a  parallelogram 
be  joined  the  area  of  the  triangle  thus  formed  is  one-eighth  that 
of  the  parallelogram. 

PROPOSITION    IV. 

399.  THEOREM.     //  a   regular  polygon  is  in- 
scribed in  a  circle  and  tangents  are  drawn  at  the 
mid  points  of  the  arcs  subtended  by  the  sides: 

I.     A  regular      circumscribed      polygon      is 
formed. 

II.  The  sides  of  the  circumscribed  polygon 
are  parellel  to  the  sides  of  the  inscribed  polygon, 
each  to  each. 


196  PLANE  GEOMETRY 

i 

III.     The  vertices  of  the  circumscribed  polygon 
lie  in  the  extended  radii  of  the  inscribed  polygon. 


A' 


C' 

Given   a  regular  inscribed   polygon   ABC ,   with 

Mf  N,  P, .  . .  .   the  mid  points  of  the  arcs  subtended  by 

the  sides  AB,  BC,  CD,   and  tangents  A'B',  B'C', 

C'D', through  the  points  M,  N,  P, 

To  Prove.   I.  A'B'C' . . .  is  regular;  II.  AB  I!  A'B', 
etc.;  III.  A'  lies  in  OA  produced,  etc. 
I.     SUG.     §  394. 

II.     SUG.     Draw   radius   OH  1  AB    and   ex- 
tend it.     "Where  does  it  meet  arc  AB*t     How  do 
AB  and  A'B'  lie  with  reference  to  OH°!     Com- 
plete the  argument. 
III.     SUG.    1.     LA=LA'.    Why? 

2.  OA  bisects   Z  A  and  OA'    bi- 
sects Z  A'.    Why?    §  393. 

3.  OEM  is  a  straight  line  LAB 
and  A'B'.     Hence  Z  MOA  =  Z  MOA' .    Why? 

4.  Hence  A'  lies  on  OA.    Why? 
Therefore 

PROPOSITION   V. 

400.     PROBLEM.    To  inscribe  a  regular  hexagon 
in  a  circle. 

Given  a  circle  with  radius  r. 
To  Inscribe  a  regular  hexagon. 


REGULAR  POLYGONS  197 

Solution.     SUG.  1.     The  radius  and  side  of  the  re- 
quired hexagon  are  equal.    Why  ? 

2.  With    the    dividers    lay    off    six 
equal  arcs.    What  is  the  length  of  the  subtended 
chord  ? 

3.  Complete  the  constructior 

PROPOSITION  VI. 

401.  THEOREM.  //  the  radius  of  a  circle  is  di- 
vided into  extreme  and  mean  ratio,  the  greater 
segment  equals  one  side  of  a  regular  decagon  in- 
scribed in  the  circle. 

Given  O  0  with  radius  0  A  divided  at  C 
into  extreme  and  mean  ratio,  OC  being 
the  greater  segment. 

To  Prove  OC   equal   to    one    side    of  a       V  / 

regular  decagon  inscrioed  in  the  circle. 

1.  Draw  the  chord  AB  equal  to  OC  and 
join  0  and  C  to  B. 

2.  It  is  to  be  shown  that  arc  AB  is  one- 
tenth  of  the  circle. 


3.  -  =       .    Why?     §327. 
OC     AC 

OA     AB 

4.  -  —  --   WViv^ 
AB      AC  * 

5.  A  OAB  <-*  A  BAG.    Why  ?    §  283. 

6.  Compare  Z  ABC  with  Z  0;  L  CBO 
with  Z  0;  L  ABO  with  Z  0. 

7.  Hence  Z  0  is  what  part  of  2  rt.  A.  ? 
Of  4  rt.  1  ? 

8.  Hence  AB  subtends  what  fractional 
part  of  the  circle  ? 

Therefore  — 


198  PLANE  GEOMETRY 

402.  COR.     A  regular  decagon  can  be  inscribed 
in  a  circle  by  dividing  the  radius  in  extreme  and 
mean  ratio  and  taking  the  greater  segment  as  the 
side  of  the  required  polygon. 

1.         Inscribe  a  regular  polygon  of  20  sides. 
J2.         Inscribe  a  regular  pentagon. 

3.  Inscribe  a  regular  polygon  of  fifteen  sides, 
called  a  penta  decagon. 

Sue.  1.     Draw  a  chord  AB  equal  to  the  radius,  and  AC 
equal  to  the  side  of  a  regular  inscribed  decagon. 

2.  What  part  of  the  circle  is  subtended  by  AB? 
by  AC? 

3.  Hence    what    part    of    the    circle    is    subtended 
by  CB? 

4.  Complete  the  construction. 

4.  What  regular  polygons  can  be  inscribed  the  construction 
of  which  can  be  based  upon  that  of  the  regular  penta-decagon? 

5.  If  the  area  of  a  regular  inscribed  triangle  is  30,  what  is 
the  area  of  the  regular  circumscribed  triangle? 

0.         If  the  area  of  an  inscribed  square  is  45,  what  is  the  area 
of  the  circumscribed  square? 

7.  What  is  the  numerical  ratio  of  the  apothem  of  a  regular 
inscribed  hexagon  to  that  of  a  regular  circumscribed  hexagon? 

PROPOSITION  VII. 

403.  PROBLEM.    To  inscribe  a  square  in  a  circle. 
Solution.     SUG.  1.     Two    vertices    must    be    the    ex- 
tremities of  a  diameter.    Why? 

2.  How  can  the  two  other  vertices 
be  located? 

3.  Draw  a  diameter  and  complete 
the  construction. 

Proof. 

8.  Inscribe  a  regular  polygon  of  8  sides;  of  16  sides;  of  32 
sides. 


R  KGTLAR  POLYGONS  199 

PROPOSITION    VIII. 

404.     THEOREM.    Regular  polygons  of  the  same 
number  of  sides  are  similar. 


A'         B' 

V- 

D' 


Given  AD  and  A'D'  ,  two  regular  polygons  of  the 
same  number  of  sides. 
To  Prove  AD  ^  A'D'. 

Proof.     SUG.  1.     What  must  be  established  to  prove 
the  similarity? 

2.     Compare    the    corresponding    angles 
A  and  A'  ,  B  and  B'  ,  etc. 


3       =1'=1'Bte-  Why? 

4.  Compare  the  ratio 

AB  A'B'      AB  BC     • 

-  with  -    —  :  —  -  with  --  . 
BC  B'C'    A'B'  B'C' 

5.  Compare 

BC        .   B'C'     BC      .;,     CD 

—  with  -  :  --  witu  --  »  etc. 

CD  C'D'    B'Cf  C'D' 

AB       BC        CD 

6.  2-==  --  —  ---  »  etc. 
A'B'     B'C'      C'D' 

Therefore— 

1.  What  is  the  locus  of  the  centers  of  circles  which  are  tan- 
cre::'t  to  a  given  line  at  a  given  point? 

2.  Two  parallel  chords  of  a  circle  are  respectively  36  and  48 
inches  long.     The  radius  is  30  inches.     Find  the  distance  between 
the  chords. 


200  PLANE  GEOMETRY 

1.  Find  the  dimensions  of  a  rectangle  which  has  a  perimeter 
of  16  in.  and  an  area  of  15  sq.  in. ;  the  area  of  one  with  a  perim- 
eter of  28  ft.  and  an  area  of  10  ft. 

2.  If  a  parallelogram  be  inscribed  in  or  circumscribed  about 
a  circle,  the  diagonals  pass  through  the  center. 

PROPOSITION    IX. 

405.  THEOREM.  The  perimeters  of  two  similar 
regular  polygons  have  the  same  ratio  as  their 
radii  and  their  apothems. 


\  __/ 


Given  AD  and  A'D'}  two  similar  regular  polygons 
with  OA  and  0'  A'  their  respective  radii,  OM  and  O'M' 
their  respective  apothems,  AB  and  A  'B'  two  homolo- 
gous sides,  and  p  and  p'  their  respective  perimeters. 

p       OA        OM 
To  Prove  —  =  ---  =  --  . 
p'     O'A'     OfMf 

Proof.     SUG.  1.     Compare   —  ;  with  -   —  .    Autli. 

p'  A'  B' 

2.     Draw  OB  and  0  'B'      Then 
A  AOB  =  A  A'O'B'.     Why? 

OA        OM       AB 


4.     Complete  the  proof. 

Therefore— 

406.  COB.  The  areas  of  two  similar  regular  poly- 
gons have  the  same  ratio  as  the  squares  of  their  radii  and 
the  squares  of  iheir  apothems. 

The  proof  similar  to  that  of  §  405  is  left  to  the  pupil. 


REGULAR  POLYGONS  201 

PROPOSITION   X. 

407.  THEOKEM.  The  area  of  a  regular  polygon 
is  equal  to  one-half  the  product  of  its  perimeter 
and  apothem. 


Given  a  regular  polygon  AD,  with  area  denoted  by 
K,  perimeter  by  p,  and  apothem  by  a. 

To  Prove  K  =  -  p  x  a. 

& 

Proof.     SUG.  1.  What  is  the  area  of   A  AOE1     Of 
BOC1 

2.  What  is  the  area  of  each  A  ? 

3.  What  is  the  sum  of  the  areas  of  all 
the  A? 

Therefore — 

In  answering  Sug.  3,  let  JL  a  be  the  unit  of  addition  and 
AB,  BC,  etc.,  be  the  respective  coefficients;  or  indicate  the  addi- 
tion and  divide  by  the  common  factor  la. 

408.  POSTULATE.  //  the  number  of  sides  of  a  reg- 
ular inscribed  polygon  be  increased  indefinitely,  th# 
apothem  approaches  indefinitely  near  the  radius  in 
length,  likewise  the  perimeter  approaches  indefinitely 
near  the  circle  and  the  area  of  the  polygon  approaches 
indefinitely  near  the  area  of  the  circle. 

A  more  exact  statement  of  the  postulate  is  as  follows : 

By  sufficiently  increasing  the  number  of  sides  of  a 

regular  inscribed  polygon  the    difference    betic&en    the 

apothem  and  the  radius  can  be  made  less  than  any  given 


202  PLANE  GEOMETRY 

number,  however  small.  The  same  is  true  of  the  perim- 
eter of  the  polygon  and  the  circle  and  the  areas  of  the 
two  figures. 

1.  What  must  be  the  nature  of  a  parallelogram  in  order  that 
a  circle  can  be  circumscribed  about  it? 

2.  [f  the  radius  of  a  circle  is  12  inches,  what  is  the  length  of 
a  side  of  an  inscribed  square?    What  is  the  length  of  the  apothem? 
What  is  the  area  of  the  square? 

3.  Find  the  area  of  the  square  in  the  preceding  example  by 
use  of  the  apothem  as  a  check  upon  the  first  computation. 

±.         If  A   and  A '    are   two   similar   polygons,   m   and  m'  two 
homologous  sides,  and  A  =  2A'f  then  m  —m'  V  2. 

5.  If  A'  is  16,  find  A.     If  A'  is  12,  find  A.     If  A  is  16,find 
A'.    If  A  is  12,  find  A1 .      Find  the  coefficient  of  W  if  a  is  i    a'  • 
if  a  is  £  a'  ;  if  a  is  3a' ;  if  a  is  KA. 

6.  If  one  acute  angle  of  a  right  triangle  is  60°,  prove  that 
the  area  of  the  equilateral  triangle  constructed  on  the  hypotenuse 
is  equal  to  the  area  of  a  rectangle  the  adjacent  sides  of  which 
are   the  two   legs   of  the  right  triangle. 

7.  If  two  triangles  have  two  sides  of  one  equal  to  two  sides 
of  the  other  respectively  and  the  included  angles  supplementary, 
they  are  equal.  A, 

The   diagonals   of    a  parallelogram    di- 
vide  it  into  four  triangles  equal  in  area.  /    / 

9.  Draw  two  concentric  regular  hexagons, 
one  being  twice  the  size  of  the  other. 

SUG.      Having    drawn    one,    with    radius    OA,    let    x  =  OA'    be 
the   unknown  radius   of  the  other.     Then  x  can  be   found 

from   the   proportion       A  AOB     =   1    =   °£  . 
A  A'O'B'  xs 

10.  If  10  is  the  length  of  the  radius  of  a  regular  hexagon, 
what  is  the  length  of  the   radius  of  a  regular  hexagon  twice  as 
large? 

11.  Divide  a  regular  hexayou   into  four  equal    parts    by    con- 
centric regular  hexngons. 


REGULAR  POLYGONS  203 

PROPOSITION  XI. 

409.  THEOEEM.  Tico  circles  have  the  same 
ratio  as  their  radii. 

Given  two  circles,  c  and  c'  with  radii  r  and  r',  respect- 
ively, p  and  p'  the  respective  perimeters  of  the  similar 
inscribed  polygons,  P2,  p/  being  the  respective  perime- 
ters of  two  similar  inscribed  polygons  of  double  the  num- 
ber of  sides,  etc. 

To  Prove 

c'     r' 

Proof.    1.     By  §405   '?*-  =  £*-  =  £*-  ......  =  *L. 

Pi        Pa'       P*  ?' 

2.  According  to  §  408  the  difference  between 
the  circles  and  the  perimeters  of  the  inscribed 
polygons  can  be  made  as  small  as  one  chooses. 
Hence  if  x  and  x'  be  any  small  numbers  there 
must  be  a  point  in  the  sequence  of  inscribed  poly- 
gons where  c  —  p  and  c'  —  p'  are  less  than  x 
and  x'  respectively.  If  these  small  differences  be 
denoted  by  y  and  yf  ,  then  c  —  p  —  y  and  c'  —  p' 

~y'-  r>    vr 

3      Hence         —        may  be  written 
r        r' 

c-j_  __  c'—y' 

r  r' 

4.     From  3  follows  -  =  --£-  & 


r     r         r       r 

5.  But  since  y  and  y'  can  be  taken  smaller 
than  any  given  number  the  expression  in  paren- 
thesis may  for  all  practicable  purposes  be  neg- 
lected, 

/>  /•» 

6.  Hence      —  —  —  • 
Therefore—  r      r 


204  PLANE  GEOMETEY 

< 

410.  COR.    I.     The  ratio  of  a  circle  to  its  diameter  is 
a  fixed  number  whatever  the  radius. 

f*  *•  O-*»  // 

Proof.     From  the  above  proposition  -=-=—=—• 

c      r      2r'     d' 

in  which  d  and  d'  are  the  respective  diameters.     From 

XI  y-,' 

this  follows     -  =  —  .    From  this  proportion  follows  the 
d        d' 

corollary.    This  ratio  is  represented  by  the  Greek  letter  TT. 

/> 

411.  COR.  II.      -  =  TT  or  c  =  Trd  =  2?rr. 

d 

412.  COR.   III.     The  area  of    two    circles    have    the 
same  ratio  as  the  squares  of  their  radii. 

Let  C  and  Cf  represent  the  two  areas.  Complete  the 
proof  in  a  manner  similar  to  that  of  §  409. 

In  this  proof  the  area  of  a  circle  can  be  approximated 
to  any  desired  degree  of  accuracy,  far  beyond  any  de- 
mands of  practical  measurement. 

413.  COR.   IV.     The  areas  of  two   circles  have   the 
same  ratio  as  Uie  squares  of  their  diameters. 

414.  COR.  Y.    Areas  of  similar  sectors  of  two  circles 
have  the  same  ratio  as  the  squares  of  their  respective 
radii  and  diameters. 

1.  Two  tangents  are  drawn  to  a  circle  of  8  in.  radius  from 
a  point  12  in.  from  the  center.     Find  the  length  of  the  chord  join- 
ing the  points  of  tangency. 

2.  The  legs  of  a  trapezoid  are  eac.h  15  inches  and  the  bases 
are  12  and  30  inches  respectively.    Find  the  area. 

3.  Two  registers  similar  in  form  are  respectively  10  and  20 
inches  in  width.     What  is  the  ratio  of  the  amounts  of  air  pass- 
ing through  them,  the  pressure  being  the  same? 

4.  Two   circular  windows  are  respectively  24  and   30  inches 
in  diameter.     What  is  the  ratio  of  their  respective  efficiencies  for 
admitting  light? 


AREA  OF  CIRCLES  205 

1.  What  is  the  locus  of  the  vertices  of  all  isosceles  triangles 
on  a  given  base? 

2.  Given  A  ABC  with  D  any  point  in  BC  extended.     Find 
a  point  E  in  AB  or  in  AB  produced  such  that  the  area  of  A  EBD 
will  equal  that  of  &ABC-.   will  be  one-half  the  area  of  &ABC: 
will  be  double  the  area  of  A  ABC :   will  have  any  given  ratio  to 
the   area   of   A  ABC. 

PROPOSITION  XII. 

415.     THEOKEM.     The  area  of  a  circle  is  equal 
to  one-half  the  product  of  its  length  and  its  radius. 


Given  a  circle  o,  with  radius  r,  length  c  and  area  A. 

To  Prove  A  =  ^  cr. 

Proof.  SUG.  1.  If  a  sequence  of  regular  polygons 
be  inscribed,  in  each  succeeding  polygon  of  the 
sequence  the  number  of  sides  being  doubled,  the 
apothem  approaches  the  radius,  the  perimeter  ap- 
proaches the  circle,  and  the  area  of  the  polygon 
approaches  that  of  the  circle.  §  408. 

2.  The  area  of  any  of  the  polygons  is 
J  ap,  a  and  p  representing  the  apothem  and  perim- 
eter respectively. 

3.  Substituting     for    these     quantities 
those  to  which  they,  by  this  process,  are  made  to 
approach,  one  obtains  A  =  J  cr. 

Therefore — 

NOTE.  A  more  rigorous  demonstration  of  the  above 
theorem  and  of  similar  theorems  which  follow  can  be 
made  by  use  of  the  theory  of  limits.  In  such  a  demon- 


206  PLANE  GEOMETRY 

* 

stration  the  possibility  of  making  the  substitution  in 
step  3  of  the  above  demonstration  would  be  considered 
and  proved  in  some  detail.  It  is  not  thought  wise  to  in- 
troduce such  considerations  at  this  point. 

416.  COR.  I.     The  area  of  a  circle  equals  nr2  or  — 

4 
Proof.    Substitute  for  c,  in  -J  cr,  its  value  2  TT  r. 

417.  COB.  II.     The  area  of  a  sector  equals  one-half 
the  product  of  its  radius  and  its  arc. 

Proof.  SUG.  1.  Prove  by  a  method  similar  to  that  of 
§  297  that  two  sectors  are  to  each  other  as  their 
angles. 

2.  Sector  of  arc  a  a 

Sector  of  arc  2-n-r      2-n-r. 

3.  But  a  sector  of  arc  2?rr  has  an  area 
of  Trr2.    Substitute  this  and  obtain  a  sector  of  arc 
a  =  |  a  X  r. 

NOTE. — Since  the  formulas  for  area  involve  the  number  TT,  their 
accuracy,  as  well  as  that  of  the  formula  for  length,  depend  upon 
the  accuracy  with  which  the  number  TT  is  computed.  The  pupil 
should  note  that  this  computation  has  not  yet  been  made.  §  419. 

1.  The  angle  of  a  sector  is  30°  and  the  radius  is  24  ft.   What 
is  the  area  of  the  sector?     Express  in  terms  of  TT. 

2.  The  area  of  a  sector  is  88  sq.  in.  and  its  angle  is  60°. 
Find  the  diameter  of  the  circle. 

3.  The  moon  has  approximately  one-fourth  the  diameter  of 
the  earth.     At  a  point  equidistant  from  them  what  is  the  ratio 
of  their  respective  "moonshine"  powers? 

4.  Two  rectangular  windows  similar  in  form  are  respectively 
2'  6"  and  3'  in  width.    What  is  the  ratio  of  the  amounts  of  light 
they  admit,  other  conditions  being  equal? 

5.  What  is  the  locus  of  a  point  equidistant   from  two   con- 
centric circles? 

6.  Cut  the  sides  of  an  equilateral  triangle  with    three  sects 
so  as  to  form  a  regular  hexagon. 


AREA  OF  CIRCLES  207 

PROPOSITION    XIII. 

418.  PROBLEM.  Given  the  radius  of  a  circle 
and  the  side  of  a  regular  inscribed  polygon,  re- 
quired to  find  the  side  of  a  regular  inscribed  poly- 
gon of  double  the  number  of  sides,  in  terms  of  the 
given  quantities. 


Given  a  circle  0  with  radius  r,  AB  a  side  of  a  regular 
inscribed  polygon,  and  AC  a  side  of  a  regular  inscribed 
polygon  of  double  the  number  of  sides. 
To  Determine  A  C  in  terms  of  AB  and  r. 

SUG.  I.    In  A  APC  express  AC  in  'terms  of  AP 
and  CP  (  §376)  and  then  in  terms  of  AB  and  CP. 

2.  Express  CP  in  terms  of  r  (i.  e.  OC) 
and  OP.    Express  OP  in  terms  of  r  (i.  e.AO)and 
AP  (  §376)  and  then  in  terms  of  r  and  AB. 

3.  Express  CP  in  terms  of  r  and  AB. 

4.  Express  AC  in  terms  of  r  and  AB. 
This  relation  when  simplified  becomes 


AC  = 

5.     If  r  be  taken  equal  to  unity,  this  be- 

comes   AC  =  ^2  -  V4-Z02 

A  partial  statement  of  the  details  of  the  above 

steps  is  as  follows  : 

AC  = 


208  PLANE  GEOMETRY 


4 

By  means  of  this  formula,  if  the  length  of  the  perim- 
eter of  any  regular  inscribed  polygon  is  known,  the 
length  of  the  perimeter  of  a  regular  inscribed  polygon 
of  double  the  number  of  sides  can  be  computed.  From 
this  result  the  perimeter  can  be  computed  in  like  manner 
for  a  polygon  the  number  of  sides  of  which  is  again  dou- 
bled. This  process  can  be  continued  indefinitely. 

PROPOSITION  XIV. 

419.  PROBLEM.  To  compute  approximately 
the  ratio  of  a  circle  to  its  diameter. 

SUG.  1.     For  convenience  the  radius  is  taken 
as  unity.    Why  may  this  be  done  ? 

2.  The  perimeter  of  a  regular  inscribed 
hexagon  is  6  which  may  be  regarded  as  a  first  ap- 
proximate value  of  the  circumference.  The  first 

t)      6 
approximation  of  TT  is  then    v*= 

3.  From  the  formula  of  §  418,  the  value 
of  sl2,  which  represents  the  length  of  a  side  of  a 
regular  inscribed  polygon  of  12  sides  is 


2- V  (4-12;  =  .  51763809 
and  consequently  p12,  or  the  perimeter,  is  12s12,  or 
6.21165708.     From  this  is  obtained  a  second  and 
closer  approximation  of 


AREA  OF  CIRCLES 


209 


TT,    .  e       -   =  6.21165708  =  3.10582854. 
d 

4.     Similarly 


-  (5.5176380)* 

and  p24  —24  x  s24. 

5.     Certain  of  the  computed  results  are 
given  in  the  following  table  : 


No. 
Sid's 

One  Side 

Perimeter 

7T 

C 
12 

24 

48 
96 
J92 

1 
V2-V3~  =  .51763809 

6. 
6.21165708 

6.26525722 

6.27870041 
6.28206396 
6.28290510 

3 

3.10582854 

3.13262861 

3.13935020 
3.14103198 
3.14145255 

V2—  V4—  (.51763809)  2=  .26105238 

^  1—  V4—  (.26105238)8  =  .13080626 

^  2—  V  4—  (.13080626)2=  .06543817 

V  2—  V  4—  (.0654381  7)  8=---  .03272346 

The  approximation  of  TT  in  common  use  is  3.1416.  For 
ordinary  work  3^  is  sufficiently  accurate. 

1.  Inscribe  a  regular  triangle.     A  regular  polygon  of  twelve 
sides.     Of  twenty-four  sides. 

2.  Measure  the  circumferences  of  several  circular  objects  as 
a  plate,  the  end  of  a  pail,  barrel,  or  stove  pipe.     Divide  the  cir- 
cumference by  the  diameter.     Average  the  several  results  to  deter- 
mine an  approximate  value  of  TT. 

3.  Compute  the  length  of  the  perimeter  of  a  24  sided  poly- 
gon, P24,  with  radius  unity  and  compare  the  result  with  that  given 
in  the  table  of  §  419. 

4  If  the  radius  is  unity,  what  is  the  length  of  one  side  of 
a  regular  inscribed  triangle? 

5.  The  radius  is  one  and  a  side  of  a  regular  inscribed  tri- 
angle is  V~3~.  Use  the  following  formula  to  find  the  side  of  a 


210  PLANE  GEOMETRY 

regular  inscribed  hexagon.    Verify   the   result  by  §  4!,S.    Formula. 

*.=  V  2-V  (4-VJ 

.* . .      A   second  method  of  solving  the  problem   of  M 

§418  is  as  follows:  /TN 

To  find  AC  in  terms  of  r  and  AB.  (      P  JA- 

Sue.     AC=-^MC  X  PC.  ic 

420.  SCHOLIUM.  Archimedes  (born  287  B.  C.)  found 
an  approximate  value  for  TT.  He  proved  that  its  value 
is  between  3^  and  3-^f .  In  modern  times  the  value 
of  TT  has  been  computed  to  a  large  number  of  decimal 
places,  two  men,  Clausen  and  Dase  independently  of 
each  other  having  computed  the  value  to  the  two-hun- 
dredth decimal  place.  Other  computers  have  given  the 
value  to  over  five  hundred  decimal  places  but  their  re- 
sults have  not  been  verified.  The  number  TT  is  incom- 
mensurable with  1,  i.  e.  it  is  neither  an  integer  or  a 
fraction,  and  hence  cannot  be  expressed  exactly  by  any 
number  of  decimal  places. 

1.  The  radius  of  a  circle  is  15  rods.  Find  its  length,  or 
circumference,  and  its  area. 

2^  With  a  tape  line  measure  the  circumference  of  a  tree  and 
compute  its  diameter. 

3.  Draw  a  circle  on  paper  or  blackboard.     Measure  its  diam- 
eter and  compute  its  length  and  its  area. 

4.  A  circular  silo  is  30  ft.  in  diameter.      How  many  squaro 
feet  in  its  floor? 

5.  Given  a  circle  with  radius  5.     What  is  its  circumference 
or  length  and  area? 

6.  The  length  of  a  circle  is  fourteen.     What  is  its  diameter 
and  its  area? 

7.  The  area  of  a  circle  is  28.     Find  the  length  and  the  diam- 
eter. 

8.  Four  6  in.  circles  are  tangent  to  each  other.     What  is  the 
area  of  the  smallest  square  that  can  enclose  thrni?     Auth. 


EXERCISES  211 

9.       "What  is  the  area  of  the  space  enclosed  by  the  circles  of 
the  preceding  exercise? 

10.  What  is  the  area  of  the  circle  externally  tangent  to   the 
above  four  circles? 

11.  What  is  the  area  of   the   circle  internally  tangent  to  the 
above  four  circles. 

12.  Three    6   in.    circles   are   tangent  externally,    what    is    the 
area  of  the  triangle  of  their  centers?     What  is  the  area  of  the 
space  included  between  the  circles? 

13.  What  is  the  length  of  the  apothem  and  what  is  the  area 
of  the  triangle  circumscribing  them? 

14.  What   is  the  area  of   the  circle   tangent   to   them   inter- 
nally? 

15.  If  the   area   included   by   three   equal    circles   tangent  to 
each  other  is  5  acres,  what  is  their  radius? 

16.  Within  a  given  circle  construct  3  equal  circles  tangent  to 
each  other  and  to  the  given  circle.     Ex.  2,  p.  37. 

17.  Construct  three  circles  tangent  to  each  other   and  exter- 
nally tangent  to  a  given  circle.     See  ex.  2,  p.  37. 

18.  Within    a   regular  triangle   construct   three   equal   circle?, 
each  tangent  to  the  two  others  and  to  two  sides  of  the  triangle. 

19.  A  circle  is  escribed  to  a  triangle  when  it  is 
tangent     to     one     side  and  to  the  two  other  sides 
produced.        A  triangle  may  have     three    escribed 
circles.        Given    a    triangle,    construct    the    three 
escribed  circles. 

Sue.     Consider  the  principle  of  inscribing  a  circle  in  a 
triangle. 

20.  Find  the  area  of  the  ring  included  between  two  concen- 
tric circles  with  radii  of  8  and  10  inches  respectively. 

21.  Find  the  length  of  a  sect  which  is  a  chord  of  one  of  two 
concentric  circles  and  a  tangent  to  the  other.     Express  the  length 
in  terms  of  the  two  radii. 

22.  Circles  are  described  upon  the  three  sides  of  a  right  tri- 
angle as  diameters.     Show  that  the  one  on  the  hypotenuse  equals 
the  sum  of  the  others. 

23.  The  diameter  of  a  circle  is  one  of  the  legs  of  an  isosceles 
triangle.     Show  that  the  base  is  bisected  by  the  circle. 


212  PLANE  GEOMETRY      . 

24.  A  three  inch  and  a  five  inch  drain  tile  unite.     What  size 
of  tile  is  necessary  to  continue  the  drain? 

25.  Three  5-in.  tiles  unite.     What  size  of  tile  should  be  used 
at  the  union?     Two  methods. 

26.  If  a  straight  line  cuts  two  concentric  circles,  the  segments 
included  between  the  circles  are  equal. 

27.  If  two    diameters  intercept  at  right  angles,   the  sum  of 
the  squares  upon  the  segments  equals  the  square  upon  the  diameter. 

28.  A  saw  log  is  15  in.  in  diameter.     What  is  the  area  of  the 
cross  section  of  the  largest  squared  timber  which  can  be  sawn  from 
this  log?     What  proportion  of  the  log  goes  into  slabs? 

29.  If  the  above  log  were  cut  into  a  timber  with  rectangular 
cross  section  eight  inches  thick,  what  would  be  the  cross   section 
area?     If  the  slabs  are  wasted,  which  of  these  two  methods  of 
sawing  is  more  profitable? 

30.  The  radius  of  a  circle  inscribed  in  a  regular  triangle  is 
one-third  its  altitude. 

31.  Show  that  seven  equal  circles  can  be  so  drawn  that  one 
of  them  is  "  internally "  tangent  to  the  six  others  and  that  each 
of  these  six  is  tangent  to  two  of  the  six. 

32.  Show  that  one  circle  can  be  so   drawn  as  to  include  the 
circles  of  the  preceding  exercise  and  be  tangent  to  six  of  them. 

33.  What  part  of  the  area  of  the  large  circle  in   above  ex. 
is  included  in  the  area  of  the  seven  equal  circles? 

34.  If  the  radius  of  the  equal  circles  is  4,  what  is  the  area 
of  that  portion  of  the  large  circle  which  is  external  to  the  small 
circles  ? 

35.  Six  dimes  can  be  placed  so  that  each  is  tangent  to  two 
others  and  also  to  a  seventh  dime.    What  is  the  ratio  of  the  diam- 
eter of  the  dime  to  that  of  the  circle  which  would  circumscribe 
them  all? 

36.  The  sides  of  two  regular  hexagons  are  3  and  5.     If  the 
area  of  the  former  is  M,  what  is  the  area  of  the  latter? 

37.  A  regular  hexagon  of  63  square  ft.  is  inscribed  in  a  cir- 
cle.    Another  is  circumscribed  about  the  circle.     What  is  the  area 
of  the  latter? 


CHAPTER  VI. 

INCOMMENSURABLE  MAGNITUDES. 

421.  In   the  discussion  of  quantities  thus  far  only 
commensurable   magnitudes  have   been   considered,   ex- 
cept in  theorems  involving  TT. 

INCOMMENSURABLE  MAGNITUDES.  (§249.)  Two  mag- 
nitudes of  the  same  kind  that  can  not  both  be  exactly 
measured  by  the  same  unit  however  small  are  incom- 
mensurable. 

Thus  V  3  and  5  are  incommensurable. 

422.  INCOMMENSURABLE  RATIO.    The  ratio  of  two  in- 
commensurable quantities  is  an  incommensurable  ratio 
or  an  incommensurable  number. 

Thus   the   ratio  y3  is   an   incommensurable   number. 

Approximations  to  the  value  of  an  incommensurable 
number  may  tag  obtained  by  neglecting  fractional  parts 
of  the  unit  used.  For  example,  the  use  of  1  as  a  unit 
gives  as  a  first  approximation  of  ^  3  the  number  1. 
The  use  of  .1  as  a  unit  gives  17.  The  use  of  .01  as  a 
unit  gives  173.  The  use  of  .00001  as  a  unit  gives 
173205.  This  approximation  may  be  carried  to  any 
desired  degree  of  accuracy  by  the  use  of  still  smaller 
units. 

An  incommensurable  number  is  neither  a  whole  num- 
ber nor  a  fraction,  for  if  it  were  either  the  original  quan 
tities  would  not  be  incommensurable  with  respect  to  each 
other. 

The  pupil  wust  not  lose  sight  of  the  fact  that  an  incom- 
mensurable number  is  as  truly  existent  as  any  other.  This  fact 


214  PLANE  GEOMETRY      . 

is  well  illustrated  by  the  consideration,  for  example,  of  the  diag- 
onal of  a  1  in.  square.  That  such  a  sect  exists  and  that  it  has 
a  length  is  not  questioned.  Geometry  proves  that  its  length 
expressed  in  inches  is  V~2T  This  is  an  incommensurable  number 
for  if  not  it  is  an  integer  or  a  fraction.  But  it  is  neither  of 
these  because  it  is  a  number  that  when  squared  gives  2.  It  is 
quite  evident  that  no  integer  when  squared  gives  2  and  also  that 
no  fraction  when  squared  will  give  an  integer. 

OL  2 

Thus  if     —  =  —  ,       a  and  &  cannot  be  incommensurable  with 
o        3 

respect  to  each  other.     This  is  seen  as  follows:     By  proportion 

^~—  —  —  .      Put    &  —  a  =  r    and    obtain    a  =  2r   and   from    this 
a  2 


The  results  in  all  practical  measurements  are  but  ap- 
proximations of  the  magnitudes  measured,  depending 
for  accuracy  upon  the  skill  of  the  operator  and  the  pre- 
cision of  the  instruments  used.  But  when  the  results 
obtained  are  within  the  degree  of  accuracy  required, 
they  are  used  as  if  they  were  the  actual  magnitudes  and 
not  approximations. 

Hence  the  previous  demonstrations  involving  incom- 
mensurable magnitudes  satisfy  all  demands  of  practical 
usage.  For  theoretical  purposes  it  is  interesting  and  im- 
portant to  know  the  absolute  truths  in  the  cases  before 
considered  and  the  discussion  of  such  requires  addi- 
tional demonstrations  of  considerable  rigor. 

423.  CONSTANT.     A  quantity  that  is  given  the  same 
value  through  a  discussion  is  a  constant. 

424.  VARIABLE.    A  quantity  which  is  given  different 
values  in  a  discussion  is  a  variable.     In  general  varia- 
bles in  changing  values  pass  through  a  succession  of  val- 
ues according  to  some  law  which  serves  to  distinguish 
them  from  other  variables. 


INCOMMENSURABLE  MAGNITUDES        215 

425.  LIMIT  OF  A  VARIABLE.  When  a  variable  by  its 
law  of  change  differs  from  a  constant  by  a  variable  quan- 
tity which  may  become  and  remain  less  than  any  as- 
signed quantity  however  small,  but  not  zero,  this  con- 
stant is  the  limit  of  the  variable. 

A  variable  may  or  may  not  attain  its  limit.  This  is 
a  matter  dependent  upon  the  law  under  which  the  varia- 
ble exists.  The  successive  approximations  of  V  2  may 
be  considered  as  the  successive  values  of  a  variable  which 
has  V  2  as  its  limit.  In  this  case  the  variable  does  not 
attain  its  limit.  In  the  case  of  the  sequence  of  regular 
inscribed  polygons  used  in  the  theorems  on  the  circle, 
the  variable  apothem,  perimeter,  and  area  do  not  at- 
tain their  limits.  On  the  other  hand  variable  segments 
between  the  sides  of  a  triangle  may  decrease  in  length 
till  at  the  vertex  they  reach  their  limit  zero ;  chords  of 
a  circle  may  increase  in  length  until  they  reach  their 
limit  the  diameter. 

If  a  point  move  along  a  sect  in  such  a  manner  as  to 
traverse  one  half  of  it  in  a  given  period  of  time,  one-half 
the  remaining  sect  in  a  second  period,  and  so  on  indef- 
initely, it  will  never  traverse  the  entire  sect.  It  is  evi- 
dent, however  that  by  continuing  the  process  for  a  suf- 
ficient number  of  periods  of  time  the  portion  still  un- 
traversed  by  the  point  will  become  and  remain  less  than 
any  given  sect,  however  small.  The  entire  sect  is  then 
the  limit  of  the  variable  distance  traversed  by  the  point. 

By  increasing  the  number  of  sides  of  an  inscribed  reg- 
ular polygon  (§  408)  indefinitely  the  perimeter  of  the 
polygon  approaches  the  circle  as  its  limit  and  the  area 
of  the  polygon  approaches  the  area  of  the  circle  as  its 
limit,  for  it  is  evident  that  the  difference  between  the 


216  PLANE  GEOMETRY 

perimeter  and  the  circle  and  the  difference  between  the 
two  areas  may,  by  this  process,  be  made  less  than  any  as- 
signed amount. 

PEOPOSITION  I. 

426.  THEOREM  OF  LIMITS.     //  two  variables  are  al- 

ways equal  and  each  lias  a,  limit,  their  limits  are  equal. 

i 

Proof.  Let  the  two  equal  variables  be  x  and  y  and 
let  their  respective  limits  be  the  constants  a  and  b.  Rep- 
resent a  —  x  by  v  and  b  —  y  by  u.  Then  by  subtrac- 
tion follows  (a  —  2c)  —  (6  —  $f-)=v  —  u.  Since  x  —  y 
this  becomes  a  —  b  =  v  —  u.  By  the  definition  of  a  limit 
(§  405)  a  and  b  are  constants  and  therefore  a  —  b  is  a 
constant.  Hence  v  —  11,  which  equals  a  —  &  is  constant. 
But  since  v  and  u  may  both  be  considered  as  small  as 
one  desires  and  are  variables  the  expression  v  —  u  can- 
not be  constant  unless  it  is  zero  in  which  case  a  —  b  is 
zero  and  a  =  b. 

427.  The  following  statements  are  here  assumed  with- 
out proof,  although  demonstrations  of  them  are  easily 
made  from  the  definitions  on  limits. 

1.  The  product  of  a  variable  by  a  constant  is  a  varia- 
ble and  its  limit  is  the  product  of  the  constant  and  MIC- 
limit  of  the  variable. 

If  the  limit  of  x  is  a,  then  the  limit  of  kx  is  ka,  k  being  a  finite 
C0n8tant' 


2.  The  quotient  of  a  variable  by  a  constant  is  a  varia- 
ble and  its  limit  is  the  quotient  of  the  limit  of  the  varia- 
ble by  the  constant. 

If  the  limit  of  x  is  a,  then  the  limit  of     ?j-    IS   T  '     fe  being  a 

K  K 

finite  constant  different  from  zero. 


INCOMMENSURABLE  MAGNITUDES        217 

PKOPOSITION  II. 

428.  THEOREM.  //  a  line  is  parallel  to  the  base 
of  a  triangle,  the  sides  are  divided  into  propor- 
tional segments. 

A 


The  following  demonstration  is  for  the  case  omitted  in 
§  251. 

Given  A  ACE  with  BD  \\  CE  and  AB  incommensur- 
able with  EC. 

AB       AD 
To  Prove 

BC       DE 

Proof.  SUG.  1.  With  some  unit  commensurable  with 
BC  divide  AB  and  BC.  A  remainder,  as  HB,  will 
occur  in  the  division  of  AB.  If  parallel  lines  are 
drawn  through  the  points  of  division  DE  will  be 
divided  into  equal  segments  and  the  line  through 
H  must  fall  somewhere  between  the  intersection 
of  the  preceding  parallel  line  and  D,  as  at  I, 
otherwise  HI  would  not  be  parallel  to  the  other 
parallel  lines. 

2.  Then  by  §  251,  --  =  --. 

BC       DE 

3.  A    sequence    of    diminishing    units 
makes  AH  a  variable  with  AB  as  its  limit.    Hence 


218  PLANE  GEOMETRY 


by  §  427  -     :  is  a  variable  with  -    -  as  its  limit. 
BC  BC 

Similarly  AI  is  a  variable  with  AD    as  its  limit 

,  AI  AD 

and.  -   -  is  a  variable  with  -    -  as  its  limit. 
DE  DE 

4.     By  §  426  the  limits  of  the  two  varia- 

AH  AI  AB    AD 

bles and are   equal  and   hence—          — 

BC          DE  BC    DE 

Therefore — 

PKOPOSITION  III. 

429.  THEOKEM.  In  the  same  or  in  equal  cir- 
cles, angles  at  the  center  have  the  same  ratio  as 
their  intercepted  arcs. 


The  following  demonstration  is  for  the  case  omitted 
in  §  297. 

Given  two  circles  0  and  0'  with  Z  AOB  incommen- 
surable with  respect  to  Z  A'O'B',  AB  and  A'B'  being 
the  respective  intercepted  arcs. 

ZO       arc  AB 

To  Prove  -r^  =    77^. 

ZO       arc  A  B 

Proof.  SUG.  1.  With  an  angle  commensurable  with 
Z  0  as  a  unit  angle  measure  the  two  angles.  In 
Z  0  there  will  be  a  remainder  as  Z  MOB  which 
is  less  than  the  unit  angle,  having  the  intercepted 
arc  MB  (the  point  M  always  falling  between  the 
end  of  the  last  unit  arc  and  point  B  ) . 

2.   By  ,  297  UOM  =A*L. 

LA'O'B'    A'B' 


INCOMMENSURABLE  MAGNITUDES        219 

3.  A    sequence    of    diminishing   units 
makes  Z  AOM  a  variable  with  Z  AOB  as  its  limit. 

Hence  by  §   427  is  a  variable  with 

—  as  its  limit.     Similarly  arc  AM  is  a  va- 
LA'O'B' 

riable  with  arc  AB  as  its  limit  and  -  is  a  va- 

A'B' 
4  B 

riable  with          -  as  its  limit. 
A'B' 

4.  By  §  426  the  limits  of  these  variable 
quotients  are  equal  and  hence 

ZO  _  Z  AOB    arc  AB 
ZO'~ZA'0'£'~arc  A'B'' 

430.     A  second  demonstration  of  §  429.     Assume  that 
Z  AOB       AB  Z  AON       AB 


^ 

<  —     -  and  take  A  so  that  -  —  (1) 

LA.'0'B'     A'B'  LA'O'B'     A'B' 

Measure  the  given  angles  by  a  unit  angle  less  than 
Z  NOB.  One  point  of  division  will  fall  then  between 

N  and  B,  say  at  M  and  by  297   y^f;  =  -7^7  •        (2) 

LA-(JB        A.  n 

From  (1)  and  (2)  follows  -  ='•*—  which  is  not 

Z  AON     AB 

true  since   Z  AOM  >   Z  AON  and  arc  AM  <  arc  AB. 
Hence  the  assumption  is  false.     Similarly  it  can  be  shown 

LAQB  .  AB 

that  -          —  '  is  not  greater  than  --- 
LA'O'B  AB' 

Therefore— 

Apply  this  method  of  proof  to  Prop.  II. 

1.  What  part  of  the  diameter  of  a  circle  is  the  apothem  of 
an  inscribed  regular  triangle?  How  can  this  conclusion  be  used 
as  a  basis  for  inscribing  a  regular  triangle? 


220  PLANE  GEOMETRY 

( 

PBOPOSITION  IV. 

4-31.     THEOREM.     Two  rectangles  having  equal 
bases  are  proportional  to  their  altitudes. 


The  following  demonstration  is  for  the  case  omitted 
in  §  357. 

Given  two  rectangles  P  and  P'  with 
equal  bases  and  altitudes  a  and  a'  in- 
commensurable with  respect  to  each 
other. 

To  Prove 

P      a 

Proof.  SUG.  1.  Measure  altitudes  a  and  -a'  with  a 
unit  commensurable  with  a'.  In  the  measure- 
ment of  a  there  will  then  remain  a  segment  c  less 
than  the  unit.  Why?  Through  the  points  of 
division  pass  lines  parallel  to  the  bases. 


2.  By  §357  Beet.         =  . 

P'       Alt.  a' 

3.  A   sequence    of    diminishing    units 
makes  rectangle  AM  a  variable  with  the  limit  P. 
Hence  by  §  427  the  ratio  of  rectangle  AM  to  P' 

p 
is  a  variable  with     —     as  its  limit.    Similarly  al- 

titude AM  is  a  variable  with  a  as  its  limit  and 
hence  the  ratio  of  altitude  AM  to  a'  is  a  variable 

with  —  as  its  limit. 
a' 


INCOMMENSURABLE  MAGNITUDES        221 

4.     By  §  426  the  limits  of  these  varia- 
bles are  equal  and  hence  — =  — 

P'     a' 

Therefore^— 

Adapt  the  demonstration  of  §  430  to  this  theorem. 

PKOPOSITION  V. 

432.  THEOREM.  Two  circles  are  to  each  other 
as  their  radii. 

This  proof  of  §  409  is  based  on  the  theory  of  limits. 

Given  two  circles  c  and  c'  with  radii  r  and  r'  respect- 
ively. 

To  Prove    -  =  -• 

c'     r' 

Proof.     SUG.  1.     Inscribe  in  the  circles  c  andV  sim- 
ilar regular  polygons  with  respective  perimeters 

v       r  T 

p  and  p' .     Then  —  =— »  and  p  =  p'  x   — 
p       r'  r' 

2.  By  increasing  the  number  of  sides 
p  becomes  a  variable  with  c  as  its  limit  and  p'  be- 
comes a  variable  with  c'  as  its  limit.     Hence  by 

§  407  the  product  p'  x  is  a  variable  with 

r' 

Cf  X;  •  I    as  its  limit. 
rf 

3.  By  §  426  the  limits  of  these  varia- 
bles are  equal  and  -c  =  c'  X  —  and  hence  —  =  —  • 

r'  c'     r' 

Therefore— 

1.  A  regular  hexagon  with  6  in.  side  is  inscribed  in  a  circle. 
Find  the  area  of  the  regular  inscribed  triangle. 

2,  If  the  angle  of  a  parallelogram  is  bisected  and  the  bisector 
extended  to  the  opposite  side  an  isosceles  triangle  is  formed. 


222  PLANE  GEOMETRY 

PROPOSITION  VI. 

433.  THEOREM.     The  areas  of  two  circles  are 
to  each  other  as  the  squares  of  their  radii.  (§412) 

Prove  this  theorem    in  a  manner  like  that  of  §  432. 

434.  A  second  proof  of  §  432.     SUG.  1.     Inscribe  in 
c  and  c'  similar  regular  polygons  with  perimeters 

TJ  T 

p  and  p'  respectively.     Then—  =— •    Why? 

p'      r 

C  )*  )' 

2.  Assume    —  >  —  and  hence   c  >  c'  X  — . 

c'      r'  r' 

Also  p  <  c  and  p'  <  c'.    §  396. 

3.  Since  in  the  sequence  of  inscribed  poly- 
gons there  is  a  point  at  which  the  inscribed  poly- 
gon  is  nearer  to  c  than  any  assigned  number, 
consider    a   polygon    q    inscribed    in    c    so    that 

T 

q>  c'  X  —-       Let  q'  be  the  similar  polygon  in- 
r' 

scribed  in  c'.  Then  —  =  —>  and  q  —  q'  X   — . 
q'      r'  r 

4.  Since  q'  <  c'  it  follows  that 

Y  T  T 

q'  x  —  <  c'  X  —  and  hence  q  <  c'  X  — 
r  r'  r' 

But  as  this  is  contrary  to  fact  the  assumption  in 

step  2  is  false. 

£ 

5.  Similarly  prove  that  cannot  be 

c 

T 

less  than  —  • 
r' 

Therefore— 

435.  Adapt,  the  demonstration  §  434  to  theorem  §  433. 


SOLID  GEOMETRY. 

CHAPTER  VII 

LINES  AND  PLANES 

436.  SOLID  GEOMETRY.     That    portion    of    Geometry 
which  treats  of  figures  the  parts  of  which  are  not  con- 
fined to  a  single  plane  (§19)  is  Solid  Geometry. 

The  resultc  of  plane  geometry  furnish  the  basis  for  investiga- 
tions in  solid  geometry,  but  it  is  to  be  remembered  that  the  state- 
ments of  plane  geometry  are  made  with  respect  to  figures  which 
are  entirely  in  one  plane  and  are  not  necessarily  true  in  solid 
geometry.  For  instance,  it  has  been  proved  in  plane  geometry 
that  only  one  perpendicular  can  be  erected  to  a  line  at  a  given 
point,  but  in  solid  geometry  many  perpendiculars  can  be  so  erected. 
This  may  be  illustrated  by  the  spokes  of  a  wheel  all  of  which  are 
perpendicular  to  its  axis.  Therefore  in  solid  geometry  the  theorems 
of  plane  geometry  must  not  be  applied  unless  the  reference  is  to 
parts  of  a  figure  all  of  which  are  in  one  plane;  but  such  theorems 
may  be  applied  first  to  one  plane,  then  to  another,  and  so  on.  The 
pupil  will  be  much  helped  in  the  study  of  solid  geometry  by  notic- 
ing that  most  of  the  theorems  are  but  extensions  or  generalizations 
of  theorems  previously  studied  in  the  plane  geometry. 

437.  The  relations  of  the  parts  of  a  figure  or  figures 
in  a  plane  are  not  changed  by  moving  the  plane  con- 
taining them  from  one  position  to  another.     §  50  (4). 

438.  A  PLANE.    A  surface  such  that  the  straight  line 
joining  any  two  of  its  points  lies  entirely  in  the  surface 
is  a  plane.     (§18.)     A  plane  is  unlimited  in  extent  and 
from  the  definition  it  follows  that  if  a  straight  line  of  a 
plane  be  indefinitely  extended,  it  can  never  leave  the 


224  SOLID  GEOMETRY 

t 

plane.     A  plane  embraces  a  line,  or  is  passed  through  a 
line,  when  the  line  lies  wholly  in  the  plane. 

439.  INTERSECTION  OF  PLANES.     That  portion  of  two 
planes  which  is  common  to  both  is  their  intersection. 

440.  PLANE  DETERMINED.    A  plane  is  determined  by 
certain  lines  or  points  when  no  other  plane  can  embrace 
those  lines  or  points  without  coinciding  with  the  first 
plane. 

441.  POSTULATE.     A  plane  can  be  revolved  about  a 
line  as  an  axis. 

Hence  it  may  be  inferred  that  a  plane  while  embrac- 
ing a  line  can  take  an  infinite  number  of  positions;  and 
that,  as  a  plane  is  unlimited  in  extent,  all  points  in  space 
can  be  embraced  by  the  plane  in  the  course  of  one  com- 
plete revolution. 

PROPOSITION   I. 

442.  THEOREM.     A  plane  is  determined— 

I.  By  a  straight  line  and  a  point  without  the 
line. 

II.  By  three  points  not  in  a  straight  line. 

III.  By  two  intersecting  straight  lines. 

IV.  By  two  parallel  straight  lines. 


I.     Given,  the  straight  line  EF  and  the  point  P  not  in 
EF. 

To  Prove  that  EF  and  P  determine  a  plane. 
Proof.    SUG.  1.     Through  line  EF  pass  a  plane  and 
revolve  it  about  EF  as  an  axis  until  it  contains 


LINES  &  PLANES  225 

the  point  P.    §  441. 

2.  How    much,    can   the    plane    be    re- 
volved either  way  about  EF   and  still   contain 
point  P?    Why?    §§4,5. 

3.  How  many  planes  can  embrace  the 
given  point  and  given  line? 

4.  •'•  EF   and  P   determine   the   plane 
MN.     §  440. 

Therefore — 


II.     Given  points  D,  E,  F  not  in  a  straight  line. 
To  Prove  that  D,  E,  F  determine  a  plane. 
Proof.     SUG.     Connect  two  of  the  points,  and  com- 
plete the  demonstration. 
Therefore- 


Ill.     Given  DE  and  FG,  two  intersecting  lines. 

To  Prove  that  DE  and  FG  determine  a  plane. 

Proof.  SUG.  1.  Since  the  lines  may  be  unlimited  in 
extent,  let  E  be  any  particular  point  in  DE  other 
than  0.  Then  FG  and  E  determine  a  plane. 
Why? 

2.  This  plane  contains  the  given  lines. 
Why? 

3.  There  is  but  one  such,  for  if  two 
planes  contained  the  given  lines  they  would  each 


226  SOLID  GEOMETRY 

contain  FG  and  E.     This  is  impossible.     Case  I. 
Therefore— 

N- 


IV.     Given  DE  and  FG,  two  parallel  lines. 

To  Prove  that  DE  and  FG  determine  a  plane. 

Proof.  SUG.  1.  By  definition  of  parallel  lines  there 
is  at  least  one  plane  containing  these  lines.  Rep- 
resent it  by  N. 

2.  If  there  is  more  than  one  plane 
through  these  lines,  there  will  be  more  than  one 
plane  containing  DE  and  F,  a  point  on  FG.  This 
is  impossible  by  Case  I. 

Therefore — 

1.  A  straight  line  can  intersect  a  plane  in  but  one  point. 
Sue.     Suppose  two  points  of  the  line  to  be  on  the  plane. 

2.  Three  straight  lines  each  intersecting  the  two  others,  but 
not  in  a  common  point,  lie  in  a  plane. 

3.  What  is  the  greatest  number  of  planes  that  may  be  deter- 
mined by  two  intersecting  lines  and  a  point?     By  three  parallel 
lines? 

4.  A   carpenter   wishing   to    determine   whether    a   board    or 
other  surface  is  a  plane,  places  a  straight  edge  (try-square)  upon 
it  in  various  positions.     What  is  the  test  of  the  straight  edge  and 
why  does  it  determine  the  question  involved? 

5.  Can  two  lines  be  so  placed  as  not  to  lie  in  one  plane? 

6.  Can  four  points  be  so  placed  as  not  to  lie  in  one  plane? 
Give  half  a  dozen  illustrations  of  mechanical  difficulties  growing 
out  of  the  answer. 

7.  Which  is  the  more  likely  to  stand  firm,  a  three  or  a  four 
legged  stool?     Why? 

443.     POSTULATE.     Two  interceding    planes    have    at 
least  two  points  in  common. 


LINES  &  PLANES  227 

PROPOSITION   II. 

444.     THEOREM.     The  intersection  of  two  plane* 
is  a  straight  line. 


V 


Given  two  intersecting  planes,  M  and  X  with  two 
points  E  and  G  in  common.  §  443. 

To  Prove  that  the  intersection  of  M  and  N  is  the 
straight  line  EG.  §  439. 

Proof.  SUG.  1.  Where  does  line  EG  lie  with  respect 
to  each  plane?  Why? 

2.  Let  0  be  any  point  in  plane  M  out- 
side of  line  EG.     Can  0  lie  in  plane  Nf    Why? 
§442. 

3.  What  are  the  only  points  common  to 
planes  M  and  X .' 

Therefore— 

1.  What  is  the  locus  of  a  point  common  to  two  planes? 

445.  THE  FOOT  OF  A  LINE.   The  point  in  which  a  line 
meets  a  plane  is  the  foot  of  the  line. 

446.  PERPENDICULAR  TO  A  PLANE.     A  line  is  perpen- 
dicular to  a  plane  when  it  is  perpendicular  to  every  line 
in  the  plane  passing  through  its  foot.    The  plane  is  then 
said  to  be  perpendicular  to  the  line. 

447.  OBLIQUE  TO  A  PLANE.    When  a  line  is  oblique  to 
one  or  more  lines  of  a  plane,  it  is  oblique  to  the  plane. 

2.  Why  is  the  crease  formed  in  folding  a  piece  of  paper  for 
an  envelope  a  straight  line? 


228 


SOLID  GEOMETRY 


PROPOSITION    III. 

448.  THEOREM.  //  a  straight  line  is  perpen- 
dicular to  two  lines  of  a  plane  at  their  point  of 
intersection,  it  is  perpendicular  to  the  plane. 


Given  EO  -L  AB  and  CD  at  0,  and  plane  M  determined 
by  AB  and  CD. 
To  Prove  EO  _L  plane  M. 
Proof.     SUG.  1.     Extend  OE  to  P,  making  OP=OE, 

and  let  OH  be  any  line  of  the  plane  through  0. 

Draw  BD  and  let  H  be  the  point  in  which  it  meets 

OH.     Connect  both  E  and  P  with  the  points  B, 

H,  and  D. 

2.  In  the  A  BEP  compare  BE  and  BP. 
In  A  DEP  compare  DE  and  DPf     §  73. 

3.  Compare  A  EBD  and  PBD; .A.  EBD 
and  PBD.    Auth. 

4.  Compare    A  EBH   and   P## ;   lines 
EH  and  Pfl".    Auth. 

5.  Compare  A  #0#  and  POfi";  1  #0# 
and  FOfl".    What  relation  does  EO  bear  to  OH ? 
Or  relate  #0  to  OH  by  §  76. 

Therefore — 


LINES  &  PLANES 


229 


449.     COR.    At  a  point  in  a  plane  only  one  perpendic- 
ular to  the  plane  can  fee  erected. 


Given  EF 1  plane  M  at  point  F. 

To  Prove  EF  the  only  perpendicular  to  M  at  F. 

Proof.  If  another  perpendicular  can  be  erected,  rep- 
resent it  by  DF.  The  lines  EF  and  DF  determine  a  plane 
(§  442)  which  will  intersect  plane  M  in  a  straight  line 
as  PF.  Then  in  this  plane,  both  EF  and  DF  are  per- 
pendicular to  PF  at  point  F.  Is  this  possible  ?  Why  ? 

Therefore — 

PEOPOSITION  IV. 

450.  THEOREM.  From  a  point  to  a  plane  there 
is  one  line  which  is  shorter  than  any  other. 


Given,  a  plane  M  and  a  point  P  without  the  plane. 
To  Prove  that  from  P  to  plane  M  there  is  one  line 
shorter  than  any  other. 

Proof.     SUG.  1.     If   there    is   not   one   shortest   line, 

there  must  be   a  group   of  equal  shortest  lines. 

Let  PE  and  PF  represent  two  such  lines.       Con- 


230 


SOLID  GEOMETRY 


nect  E  and  F  and  let  0  be  the  mid-point  of  EF: 
Join  P  and  0. 

2.     With  respect  to  length,  compare  PO 
with  PE  or  PF. 
Therefore— 

451.  COR.  I.     The  shortest 
line  from  a  point  to  a  plane  is 
the  perpendicular  from   that 
point  to  the  plane. 

Given  PO  the  shortest  line 
from  point  P  to  the  plane  M. 

To  Prove  that  PO  is  per- 
pendicular to  plane  M. 

Proof.  Through  the  foot  of  PO  draw  in  plane  M  any 
two  lines  as  EF  and  GH.  Then  PO  1  EF  and  PO  1  GH. 
Why? 

/.  PO  1  plane  M.    Why? 

Therefore — 

452.  COR.  II.     From  a  point 
without  a  plane  only  one  perpen- 
dicular can  be  dropped  to   the 
plane. 

Given  PO   1   plane   N   from 
point  P  without  the  plane. 

Proof.  If  there  is  a  second  perpendicular  from  P  to 
N,  represent  it  by  PG.  Then  the  plane  PGO  (§  442) 
intersects  plane  N  in  line  OG.  Why? 

What  relation  does  PG  bear  to  line  OG1  Why? 
What  relation  does  PG  bear  to  plane  N  ?  Why  ? 

Therefore— 

453.  DISTANCE  FROM  A  POINT  TO    A    PLANE.     The 
length  of  the  perpendicular  from  a  point  to  a  plane  is 
the  distance  from  the  point  to  the  plane. 


LINES  &  PLANES 


PBOPOSITION   V. 


454.  THEOREM.     All  perpendiculars  to  a  line  at 
a  given  point  lie  in  one  plane  perpendicular  to  tlie 
line  at  that  point. 

Given  plane  M  1  PO  at  0  and  line  OH  any  line  l  PO 
at  0. 

To  Prove  that  OH  lies  in  plane  M. 

Proof.    SUG.  1.     The  plane  N  determined  by  PO  and 

OH   will   intersect   plane  M   in   a   straight   line 

through  0,  as  OG.    Why? 

2.  POA.OG.    Why? 

3.  OG  and  OH  both  lie  in  plane  N  and 
must  therefore  coincide.    Why? 

4.  •'•     OH  must  lie  in  plane  M.    Why? 

5.  As  Ofl"  was  any  line  perpendicular 
to  PO  at  0,  the  same  is  true  of  all  such  lines. 

Therefore — 

455.  COB.  I.     Through  a  point  in  a  line  only 
plane  can  be  erected  perpendicular  to  the  line. 

Given  a  line  a  and  a  plane  M 1  to  a  at  point  0. 

To  Prove  that  M  is  the  only  plane  1  a  through  O. 

Proof.  Suppose  that  there  is  a 
second  plane,  as  N,  -L  a  at  0,  and 
let  b  be  the  intersection  of  planes 
M  and  N.  Pass  a  plane  P  through 
line  a  which  does  not  contain  line 
b.  This  plane  will  cut  M  and  A7  in 


one 


232 


SOLID  GEOMETRY 


two  distinct  lines,  each  perpendicular  to  a.    But  this,  by 
§  454,  is  impossible.    Why? 

456.     COR.  II.    From  a  point  without  a  line  only  one 
plane  can  be  drawn  perpendicular  to  the  line. 


Given  point  P  not  on  line  a  and  plane  M  through  P 
and  1  a. 

To  Prove  that  M  is  the  only  plane  through  P  and  1  a. 

Proof.  Let  M  meet  line  a  in  point  A.  If  there  is  a, 
second  such  plane,  as  N,  let  it  meet  a  in  point  B.  Pass 
a  plane  through  P  and  line  AB  or  a.  This  plane  will 
cut  M  and  N,  each  in  a  straight  line.  Complete  the 
demonstration. 

1.  Erect  a  plane  perpendicular  to   a  line  at   a  given  point. 
§454. 

2.  How  can  a  carpenter  erect  a  studding  perpendicular  to  a 
floor  with  his  square? 

3.  Set  up  in  the   school  room   a   pole   perpendicular   to   the 
floor,  using  a  right  angle  (a  carpenter's  square  or  a  book  cover). 
§448. 

4.  How  many  braces   are  required   to  hold  the  pole   of  the 
preceding  exercise  in  position? 

5.  Construct   a  plane  perpendicular  to  a   given  line  from  a 
point  without  the  line.     §  448. 

6.  If  a  plane    is  perpendicular  to  a 
sect  at  its  mid-point  every  point   in   the 
plane  is  equidistant  from  the  extremities 
of  the  sect  and  every  point   outside   the 
plane  is  not  equidistant  from  the  extrem- 
ities. 


LINES  &  PLANES 


233 


Sue.   Let  X  be  any  point  in  plane  M,  which  is  perpen- 
dicular to  sect  DE  at  its  mid  point  0.     Prove  DX  =  EX. 

PROPOSITION  VI. 

457.     PROBLEM.    To  drop  a  perpendicular  from 
a  point  to  a  plane. 


Given  a  plane  M  and  point  P  not  on  M. 

To  Construct  a  line  PO  from  PA.M. 

Construction.     SUG.  1.     Draw  any  line  EF  in  plane 

M  and  drop  a  perpendicular,  PH,  to  EF  in  the 

plane  of  EF  and  P.    Auth. 

2.  Through  H  in  plane  M  draw 
HG1.EF.    Auth. 

3.  From     P     drop     POA.HG. 
Auth. 

4.  To   prove   PO  1  M,   draw   a 
line  from  0  to  1£,  any  point  in  EF  except  H,  and 
join  P  and  J£. 

5.  ~PH2-OH2=P02.     Auth. 

6.  EH2+On2=OE2.     Auth. 

7.  Adding,  #P8  =  P08  +0#2- 
S.     •'•  PO±OE. 

9.     •*•  POlM.     Why? 

1 .        What  is  the  locus  of  points  in  space  equidistant  from  two 
given  points? 


2;U  SOLID  GEOMETRY 

( 

458.  Locus  IN  SPACE.    The  locus  of  a  point  in  space 
satisfying  certain  given  conditions  consist  of  those  geo- 
metric figures  in  space  to  which  the  point  is  limited  and 
every  point  of  which  satisfies  the  conditions.  §§  164-165. 

PROPOSITION  VII. 

459.  PROBLEM.     To  erect  a  perpendicular  to  a 
plane  at  a  given  point  in  the  plane. 


Given  plane  M  and  point  0  in  M. 

To  Construct  OP  LM  at  0. 

Construction.     Sue.  1.     In  plane  M  draw  a  line  a 

and  then  two  lines  c  and  d  each  perpendicular  to 

a  at  0. 

2.  What  relation  does  the  plane 
of  c  and  d  bear  to  a  ? 

3.  This  plane  cuts  M  in  line  I). 
Why? 

4.  In  this  new  plane  draw  OP 
1  1).    Auth.    How  is  OP  related  to  af    Why  ? 

5.  How  is  OP  related  to  plane 
M?    Why? 

Compare  this  construction  with  ex.  3,  p.  232. 

1.        Another    demonstration     for    §  448.     Draw    from    D    a 
line  DB  terminated  in  line  AB  and  bisected  by  OH  at  H.     Auth. 
(Use  the  fig.  of  §44.8.)_ 
EBZ+EI)2=  2EH*  +  2  #£Ta   QBz+f)J)*  =  20H*+  2  BE2  §  345 

Subtract  and  reduce.     Apply  §  311. 


LINES  &  PLANES 


235 


PROPOSITION   VIII. 

460.  THEOREM.  If , from  the  foot  of  a  perpen- 
dicular to  a  plane,  a  line  is  drawr,  perpendicular 
to  any  given  line  in  the  plane  and,  from  this  point 
of  intersection,  a  line  is  drawn  to  any  point  of 
the  perpendicular  to  the  plane,  the  last  line  is 
perpendicular  to  the  given  line  in  the  plane. 


Given  line  EG  1  plane  M  and  CD  any  line  in  M,  with 
EO  1  CD  and  F  any  point  in  EG. 
To  Prove  OF  1  CD. 

Proof.     SUG.  1.     On  line   CD  take  OD  =  OC.     Join 
F  with  C,  D  and  0,  and  E  with  C  and  D. 

2.  In   &ECD  compare  EC  and   ED. 
Autli. 

3.  Compare  FC  and  FD.     §  65. 

4.  What    relation    does    OF   bear 
76. 


to 


CD? 
Therefore — 

1.  In    the    fig.    of    §    460,    given    EG  L  EO,    EO  LCD,    and 
OF  LCD   at   0,  F  being  in  EG.     Prove  EG  1  plane  M   of  EO 
and  CZ>. 

SUG.     EF*  =  OF^  —  £02  and  EC*  =  0(72  +  0#2. 
Add  and  complete  the  demonstration. 

2.  Show  that  by  marking  at  right  angles  to  the  edges  of  a 
stick   of   timber  with   a  carpenter's   square   continuously   the   line 
will  end  at  its  starting  point. 


236  SOLID  GEOMETRY 

PROPOSITION  IX. 

461.  THEOREM.  //  one  of  two  parallel  lines  is 
perpendicular  to  a  plane,  the  other  is  also  perpen- 
dicular to  the  plane. 

H  D 


Given  GH  II  CD  and  meeting  plane  M  in  G  and  C  re- 
spectively, with  CD  j_  M. 

To  Prove  GHi_M. 

Proof.  SUG.  1.  GH  and  CD  determine  a  plane 
which  meets  plane  M  in  GC.  Why  ?  Join  G  witk 
Oy  any  point  in  CD.  Draw  EF  in  plane  3f  and 


2.  What  relation  does  EF  sustain  to 
GO?    (§460.)    ToGCf    To  plane  #C7    §448. 

3.  What   relation  does   EF  sustain  to 
GH,  or  GH  to  #jF?     £#  to  GC  f     §  90.     £H  to 
plane  M?     Why? 

PROPOSITION  X. 

462.     THEOREM.     Two    Zmes    perpendicular    to 
the  same  plane  are  parallel. 


Given   two    lines    a    and    &,    each    perpendicular   to 
plane  M. 
To  Prove  a  II  b. 


LINES  &  PLANES  237 

Proof.     SUG.  1.     Suppose  a  is  not  parallel  to  b  and 
through  a  point  0  on  a  draw  a  line  c  I!  6. 

2.  What  relation  does  c  bear  to  plane 
Mt    Why?     §  461. 

3.  Complete  the  demonstration. 
Therefore— 

PROPOSITION  XL 

463.  THEOREM.  Two  straight  lines  each  par- 
allel to  a  third  straight  line  are  parallel  to  each 
other. 


7 


Given  a  II  c  and  b  II  c. 

To  Prove  a  II  6. 

Proof.     SUG.  1.     Construct  plane  M  ±.c.     Auth. 

2.  What  relation  do  a  and  b  bear  to 
Mf    Auth. 

3.  What  relation  do  a  and  b  bear  to 
each  other?     Auth. 

Therefore — 

].  To  draw  a  straight  line  which  shall  intersect  each  of  two 
lines  not  in  the  same  plane  and  shall  pass  through  a  point  not  in 
either  line. 

SUG.      Pass    a    plane    through    the    given 
point  P  and  one  line  a.     Also  pass  a  plane  9 

through   P   and   the   second   line    ft.      These  ,s 

planes  intersect  in  a  line  x  through  P. 
Why?  This  line  x  also  meets  both  a  and  &. 
Why? 

464.  PROJECTION  OF  A  POINT  ON  A  PLANE.  The  foot 
of  the  perpendicular  from  a  point  to  a  plane  is  the  pro- 
jection of  the  point  on  the  plane. 


238  SOLID  GEOMETRY 

e 

465.  PROJECTION   OF  A  LINE  UPON  A  PLANE.     The 
locus  of  the  projections  of  the  points  of  a  line  upon  a 
plane  is  the  projection*  of  the  line  upon  the  plane. 

PROPOSITION  XII. 

466.  THEOREM.     The  projection  of  a  straight 
line  upon  a  plane  is  a  straight  line. 


Given  a  plane  M  and  a  straight  line  EF  not  in  M.  with 

E'  and  F'  the  projections  of  E  and  F  respectively  upon  M. 

To  Prove  the  straight  line  E' F'  the  projection  of  EF. 

Proof.     SUG.  1.     It  is  necessary  and  sufficient  to  show 

that  the  projection  P'  of  any  third  point  in  EF 

lies  in  E'F'. 

2.  Since  PP' ,  EE' ,  FF'  are  each  l_M, 
they  are  parallel.    Auth. 

3.  EE'  and  FF'  determine  a  plane,  AT, 
cutting  M  in  line  E'F'.    Why? 

4.  In  plane  N,  it  is  possible  to  draw  a 
line  through  P  which  is  parallel  to  FF'  and  hence 
1  M,  but  this  line  must  coincide  with  PP',    Why? 

5.  Since  PP'   lies  in  N,  its  foot  in  M 
must  lie  in  E'F' , 

Therefore— 

1.  The  area  of  a  circle  is  20  sq.  in.     What  is  the  area  of  a 
circle  of  double  the  radius?       Of  ^   the  radius?     Of  m  times  the 

radius?      Of    —     times  the  radius? 
m 

2.  If  a  is  the  radius  of  a  circle  with  an  area  of  40  sq.  in., 
what  is  the  radius  of  a  circle  of  80  sq.  in.?     Of  10  sq.  in.?     Of 
ICO  sq.  in.? 


LINES  &  PLANES  239 

PROPOSITION  XIII. 

467.     THEOREM.     //  from  a  point  to  a  plane  the 
perpendicular  and  oblique  lines  be  drawn 

I.     The   perpendicular   is   the   shortest   line 
from  the  point  to  the  plane; 

II.     Oblique  lines    having    equal    projections 
upon  the  plane  are  equal; 

III.  //  two  oblique  lines  are  equal,  their  pro- 
jections upon  the  plane  are  equal; 

IV.  Of  two  unequal  oblique  lines,  the  line  hav- 
ing the  greater  projection  is  the  greater; 

V.     The  greater  of  two  oblique  lines  has  the 
greater  projection  upon  the  plane. 


I.  Given  PO  1  plane  M. 

To  Prove  PO   shorter  than  any  other  line   from   P 
to  .V. 

Proof.     SUG.     Compare  A  POG  and  PGO. 

II.  Given  PG  and  PE  oblique  to  M  with  their  re- 
spective projections  OG  and  OE  equal. 

To  Prove  PG  —  PE. 

Proof.     The  demonstration  is  left  to  the  pupil. 

III.  Given  PG  =  PE,  with  OG  and  OE  their  re?pec- 
tive  projections. 

To  Prove  OG  =  OE. 

Proof.    The  demonstration  is  left  to  the  pupil. 


240  SOLID  GEOMETRY 

IV.  Given  obliques  lines  PE  and  PF,  with   OE  and 
OF  their  repective  projections,  and  OE  >  OF. 

To  Prove  PE  >  PF. 

Proof.    SUG.  1.  On  OE  take  021  =  0^  and  draw  PH. 
2.     Compare   PE  with   P#;  P#  with 
PF.     Complete  the  demonstration. 

V.  Given  PE  >  PF. 
To  Prove  OE  >  OF. 

Proof.     SUG.     Use  the  indirect  method. 
Therefore— 

1,  What  is  the  locus  of  the  foot  of  an  oblique  line  in  a  plane 
when  the  oblique  line  is  revolved  with  one  end  not  in  the  plane 
stationary? 

2,  An  oblique  to  a  plane  intersects  its  own  projection  upon 
the  plane. 

3,  The  locus  of  a  point  in  space  equidistant  from  all  points 
in  a  circle  is  a  straight  line  through  the  center  of  the  circle  and 
perpendicular  to  its  plane. 

4,  A  ceiling  is  8  ft.  high.     A  pole  10  ft.  long  is  held  at  a 
given  point  on  the  ceiling  and  revolved  with  the  lower  end  touch- 
ing the  floor.     Find  the  diameter  of  the  circle  described. 

5,  How  could  the  pole  in  the  preceding  exercise  be  used  to 
find  that  point  on  the  floor  which  is  directly  under  the  point  on 
the  ceiling? 

6,  Show  that  the  angle  that  the  pole  makes  with  the  perpen- 
dicular is  constant  as  the  pole  revolves. 

7,  A  horse  is  tied  with  a  rope  100  ft.  long  to  a  point  on  a 
vertical  pole  25  ft.  from  the  ground.     What  is  the  shape  of  his 
pasture  and  how  far  can  he  get  from  the  foot  of  the  pole? 

468.  LINE  AND  PLANE  PARALLEL.    A  straight  line  and 
a  plane  are  parallel  if  they  can  never  meet,  however  far 
they  may  be  extended. 

469.  PARALLEL  PLANES.     Two  planes  which  cannot 
meet,  however  far  they  may  be  extended,  are  parallel 
planes. 


LINES  &  PLANES  241 

PROPOSITION    XIV. 

470.     THEOREM.     A  straight  line  is  parallel  to  a 
plane  if  it  is  parallel  to  a  line  in  the  plane. 


Given  line  a  outside  plane  M  and  parallel  to  line  b  in 
plane  M . 

To  Prove  a  II  M. 

Proof.     SUG.  1.     a  and  b  determine  a  plane  N.  Why? 
§  442. 

2.  What  is  the  intersection  of  M  and 
Nf    Why?    §  444. 

3.  If  a  and  M  have  a  point  in  common 
it  must  lie  in  b.    Why?    §444. 

4.  This   is   impossible.     Why  ?      §  438. 
Therefore— 

471.  COR.  I.  A  plane  may  be  passed 
through  one  of  two  non-intersecting 
lines  parallel  to  the  other. 

Given  two  non-intersecting  lines  a  and  b. 
To  Prove  that  a  plane,  M,  can  be  passed  through  b 
parallel  to  a. 

Proof.     SUG.     From  any  point  in  b  draw  the  line  c 

parallel  to  a.     Then  plane  M  determined  by    b 

and  c  is  parallel  to    a.    Why? 

1.  How  many  such  planes    (§471)   are  there? 

2.  Jf  a   line  is  parallel  to  a  plane,  it  is  parallel  to  its  own 
projection  upon  the  plane. 


242  SOLID  GEOMETRY 

472.     COR.  II.    Through  a  point  not « 
on  either  of  two  lines  a  plane  can  bo 
passed  parallel  to  both  the  lines.  A X~~- 


p. 

x' 


Given  lines  a  and  b  and  point  P. 

To  Prove  that  a  plane  M  parallel  to  a  and  to  b  can 
be  passed  through  P. 

Proof.     Through  P  draw  two  lines,  c  and  d,  parallel 
to  a  and  b  respectively.     Complete  the  demonstration. 
PEOPOSITION  XV. 

473.  THEOEEM.  //  a  line  is  parallel  to  a  plane  f 
any  plane  that  embraces  the  line  and  intersects 
the  plane  intersects  it  in  a  line  parallel  to  the 
given  line. 

N <— 


Given  line  c  parallel  to  plane  M,  and  plane  N  through 
c  and  intersecting  M  in  line  b. 
To  Prove  b  I!  c. 

Proof.     SUG.  1.     If  c  and  b  are  not  parallel  they  will 
meet  at  some  point  as  0.    Why?     This  is  impos- 
sible.    Why? 
Therefore— 

1.  Through  a  given  point  in  space  one  and  only  one  line  can 
be  drawn  parallel  to  a  given  line.     §  87. 

2.  Through  a  given  point  in  space  any  number  of  lines  can 
be  drawn  parallel  to  a  given  plane. 

3.  All  lines  through  a  given  point  and  parallel  to  a  given 
plane  lie  in  one  plane. 

SUG.     Drop  a  1  from  the  given  point  to  the  given  plane. 

4.  Through    a    given    point    only   one    plane    can    be    passed 
parallel  to  a  given  plane. 


LINES  &  PLANES  243 

PROPOSITION   XVI. 

474.  THEOREM.     Planes  perpendicular   to   the 
same  straight  line  are  par  alb  I.  \ 

Given  two  planes  M  and  A"  each       f~~             ~~}1 
perpendicular  to  line  a.  /     _J / 

To  Prove  M  I!  A7. 

Proof.     SUG.     Use  an  indirect  proof      /~~ 
based  on  §  450.  /_ 

Therefore— 

PROPOSITION   XVII. 

475.  THEOREM.     //    two   intersecting   lines    in 
one  plane  are  each  parallel  to  a  second  plane,  the 
planes  are  parallel. 


7 


7 


Given  two  lines  a  and  6  in  plane  M,  meeting  in  O, 
each  line  parallel  to  plane    AT. 
To  Prove  M  II  N. 

Proof.  SUG.  1.  At  0,  the  common  point  of  a  and  1), 
erect  a  1  to  plane  M  and  extend  it  to  meet  plane 
N  at  0'.  Line  00'  is  l  to  a  and  to  6.  Why  ? 

2.  Lines  a  and  00'  determine  a  plane 
which  intersects  N  in  a  line  as  a'   through  0' . 
Likewise  b  and  00'  determine  a  plane  which  in- 
tersects A7  in  &'.    Auth. 

3.  What   relation  does  a'   bear  to   «? 
?/  to  1}  ?    Auth. 


244  SOLID  GEOMETRY 

4.  What  relation   does  00'  bear  to  a' 
and  to  &'?     Auth. 

5.  What    relation    does    00'    bear    to 
plane  N?    Auth. 

6.  What  relation  does  N  bear  to  Mf 
Auth. 

Therefore — 

PROPOSITION  XVIII. 

476.     A  line  perpendicular  to  one  of  two  par- 
allel planes  is  perpendicular  to  the  other. 


Given  plane  M  II  plane  N  and  line  a  1 M  at  0. 

To  Prove  a_LN. 

Proof.  SUG.  1.  Through  any  point  in  plane  M  draw 
two  lines  as  b  and  c.  Through  a  and  &  and 
through  a  and  c  pass  planes  and  let  these  two 
planes  intersect  plane  N  in  lines  b'  and  c'  re- 
spectively. 

2.     What   relation   does   &'    bear   to   b? 
c'  to  c?    Auth. 

3.  What  relation  does  a  bear  to  b  and 
to  c?    To  b'  and  to  c'  ?    Auth. 

4.  What   relation    does   a    hoar   to   Nf 
Auth. 

Therefore— 

1.  Two  planes  parallel  to  the  same  plane  are  parallel  to  each 
other.    §§  474,  476. 

2.  What  is  the  locus  of  the   foot  of   an   oblique  line   10   in. 
long  drawn   to   a  plane  from  a  point  eight  in.   from   the  plane? 
Compute  the  area  of  the  figure  bounded  by  the  locus. 


LINES  &  PLANES  245 

477.  DISTANCE   BETWEEN    PARALLEL    PLANES.      The 
length   of   the    sect   intercepted   between    two    parallel 
planes  and  perpendicular  to  both  is   the  distance   be- 
tween the  parallel  planes. 

PROPOSITION  XIX. 

478.  THEOREM.     //  two  parallel  planes  are  cut 
by  a  third  plane  the  intersections  are  parallel. 


Given  two  parallel  planes  M  and  N,  cut  by  a  third 
plane  P  in  lines  m  and  n  respectively. 
To  Prove  m  II  n. 
Proof.     SUG.     Show  that  m  and  n  lie  in  the  same 

plane  and  cannot  meet. 
Therefore — 

479.  COR.    I.     Parallel    lines    intercepted     between 
parallel  planes  are  equal. 

SUG.     Assume  lines  a  and  &   in   §   478  to  be 
parallel.     Prove  them  equal. 

480.  COR.  II.     Parallel  planes  are  everywhere  equi- 
distant. 

1.  A   straight   line  and   a   plane   both  perpendicular   to    the 
same  straight  line  are  parallel. 

2.  What  is  the  locus  of   a  line  through  a  given  point  and 
parallel  to  a  given  plane? 

3.  If  a  straight  line  and  a  plane  are  parallel,  any  line  par- 
allel to  the  given  line  is  parallel  to  the  plane  also. 


246  SOLID  GEOMETRY 

t 
PROPOSITION   XX. 

481.  THEOREM.  //  two  angles,  not  in  the  same 
plane,  have  their  respective  sides  parallel  and  ex- 
tending in  the  same  directions  from  the  vertices, 
the  angles  are  equal. 


Given  Z  BAG  in  plane  M  and  Z  B'A'C'  in  plane  N, 
with  A#IIA'l?'   and  ACllA'C',  the  respective  direc- 
tions  from  A  and  A '  being  the  same. 
To  Prove  Z  BAC=  Z  £'A'C'. 

Proof.  SUG.  1.  Take  points  B,  B',  C,  C'  so  that 
AB  =  A'B'  and  AC  =  A'C'.  Connect  A  and 
A',  B  and  5',  C  and  C',  5  and  C,  B'  and  C". 

2.  Compare  AA'   and  BB\  A  A'   and 
CC',  BB'  and  CC". 

3.  Compare    BC    and    £'C";    ABAC 
and  A  B'A'C1 ;  Z  A  and  Z  A'.     Auth. 

Therefore— 

1.  What  is  the  locus  of  a  point  at  a  given  distance  from  a 
given  plane? 

2.  What  is  the  locus  of  a  point  equidistant  from  two  given 
points  and  also  at  a  given  distance  from  a  given  plane?     When 
is  there  no  solution? 

3.  From  A,  the   mid-point   of  one  side   of  a  parallelogram 
draw  lines   dividing  each  of  the  adjacent  sides  into   three  equal 
parts    and    the    opposite    side    into    six    equal   parts.      Prove    that 
each  of  the   twelve  triangles   thus   formed  equals   one  twelfth  of 
the  parallelogram. 


LINES  &  PLANES  247 

PROPOSITION   XXI. 

482.  THEOREM.  //  three  parallel  planes  inter- 
sect two  straight  lines,  the  corresponding  seg- 
ments are  proportional. 


Given  three  parallel  planes  M,  N,  P  intersecting  the 
two  lines  a  and  a'  in  the  points  D,  E,  F  and  D',  E'  ,  F' 

respectively. 

_    _  DE     D'E' 

To  Prove      —  =  — 

EF     E'F' 

Proof.     SUG.  1.     Join  D  and  F'  and  let  G  be  the  in- 
tersection of  this  line  with  plane  N. 

2.  Plane  DD'F'    intersects   planes   M 
and  N  in  what   lines?     Plane   DFF'    intersects 
planes  Ar  and  P  in  what  lines  ? 

3.  What  relation  does  EG  bear  to  FF'  ? 
E'GtoDD'1    Why? 

.       DE    DG   D'E' 

4.  Compare  the  ratios 


§251. 

5.     Complete  the  demonstration. 
Therefore— 

1,  If  two   parallel  planes   intersect   two   parallel   planes    the 
four  lines  of  intersection  are  parallel. 

2.  What  is  the  locus  of  a  point   in   space  equidistant   from 
two  parallel  planes? 


248  SOLID  GEOMETRY 

1.  Find  a  point  X  equidistant  from  two  given  points,  equi- 
distant from  two  other  given  points,  and  in  a  given  plane.     Is  the 
problem    ever   impossible?      Will    any   arrangement    of   the    given 
parts  permit  an  unlimited  number  of  solutions'? 

2.  Another  demonstration  for  461.     Given  =  AB  \\  A'B'  and 
A'B'l.  plane  M. 

To  PROVE  AB  1  plane  M. 

SUG.  Draw  in  plane  M  any  two  lines  through  B'  as 
B'C'  andB'D'.  Through  B  draw  BC  and  BD  parallel 
to  B'C'  and  B'D'  respectively.  &.  A. 

A' B'C'  and  A' B'D'  are  right  angles. 
Why?  They  are  equal  to  ^  AB 0  and 
ABD  respectively.  Why!  Therefore 
ABLM.  Why? 

3.  Find  a  point  X  equidistant  from  two  given  points,  equi- 
distant from  two  given  planes,  and  at  a  given  distance  from  a 
third  plane.     Discuss  all  possibilities. 

4.  What  is  the  locus  of  a  point  equidistant  from  two  given 
points  A  arid  B  and  also  from  two  given  points  C  and  D1    Discuss 
all  possibilities  due  to  the  different  positions  of  the  pairs  of  points. 

483.  DIHEDRAL  ANGLE.     riwo  planes  which  meet  or 
intersect  form  a  dihedral  angle.    The  two  planes  M  and 
N  meeting  in  the  line  ED  form  a  dihedral 

angle.     The  intersecting  planes,  M  and  A7, 

are  the  faces  of  the  angle  and  the  line  of 

intersection,  ED,  is  the  edge.     A  dihedral 

angle  is  read  by  reading  in  order  one  face, 

the  edge,  and  the  second  face.    When  but 

one  dihedral  angle  is  formed  at  an  edge, 

the   angle   may   be   read   by   naming   the 

edge,  as  dihedral  angle  ED.     A  dihedral  angle  is  often 

called  a  dihedral. 

484.  PLANE  ANGLE  OF  A  DIHEDRAL.    An  angle  formed 
by  two  lines,  one  in  each  face  of  a  dihedral,  perpendicu- 
lar to  the  edge  at  the  same  point  is  the  plane  angle  of 
the  dihedral    Lines  FO  and  HO  lying  in  the  faces  M 


DIHEDRAL  ANGLES 


249 


and  AT  respectively  and  each  perpendicular  to  KI)  at  O 
is  a  plane  angle  of  the  dihedral  ED. 

485.  COR.  I.     The  plane  angle  of  a  dihe- 
dral is  the  same  size  from  whatever  point 
of  the  edge  it  is  drawn. 

Given    dihedral   EE'  with   plane   angles  E> 
GEH   andG'E'H'. 
To  Prove  L  GEH=  L  G'E'H'. 
Proof.     Left  to  the  pupil.     §  481. 

486.  COR.  II.     The  angle  formed  by  the  intersections 
of  the  faces  of  a  dihedral  with  a  plane  perpendicular  to 
the  edge  is  a  plane  angle  of  the  dihedral. 

Proof  left  to  the  pupil.    §  484. 

487.  EQUAL   DIHEDRALS.     Two   dihedral   angles   are 
equal  when  they  can  be  made  to  coincide. 

The  magnitude  of  a  dihedral  does  not  depend  upon  the  extent 
of  its  faces.  If  a  plane  be  made  to  revolve  about  the  edge  as  an 
axis  from  the  position  of  one  face  to  the  position  of  the  other 
face,  it  revolves  or  turns  through  the  dihedral  angle  and  the 
greater  the  amount  of  the  turning  the  greater  the  angle. 

1.  A   and  B  are  two  points  equally  distant     from  a   plane 
and  upon  the  same  side  of  it;   prove  that  line  AB  is  parallel  to 
the  plane. 

2.  Determine  a  point  E  in  a  plane  such  that  the  difference 
between   its    distances   from  two    given   points    on   opposite   sides 
of  the  plane  is  a  maximum. 

Sue.    Drop  a  perpendicular  to  the 
plane  from  one  of  the    points  as  H 
and  extend  it  to  G,  an  equal  distance 
on  the  other  side  of  the  plane.  Connect 
G  with  the  second  point  F,  this  line 
meeting  the  plane  in  E.     Prove  E  to 
be  the  required  point,  as  follows : 
EU  —  EF  =  FG.      Take    any    other   point    as    E'    in    the    given 
plane,  join  it  to  F  and  H  and  prove  FG  >  E'H  —  E'F. 


250 


SOLID  GEOMETRY 


1.  Determine  a  point  E  in  a  plane  such  that  the  sum  of 
its  distances  from  two  points  on  the  same  side  of  the  plane  is 
the  minimum. 

Sue.    Drop  a  perpendicular  to  the  plane  from  one  of  the 
points  as  H  and  extend  it  to   G,  an 
equal  distance  beyond  the  plane.   Join          H 
G   to    the    second   point    F,    this   line          j^s^;  -.„_/% 
meeting  the  plane  in  E.     Prove  E  to 
be   the   required   point.      To    do    this, 
take   any    other   point   as  E'    in   the          «  //-" 
given   plane    and   prove   HE'  +  E'F        c 
>  HE  +  EF. 

PEOPOSITION   XXII. 

488.     THEOREM.     Two  dihedral  angles  are  equal 
if  their  plane  angles  are  equal. 


H' 

Given  two  dihedrals  GH  and  <?'J§P,  with  equal  plane 
angles  O  and  0'  respectively. 

To  Prove  dihedrals  GH  and  G'H'  equal. 
Proof.     SUG.  1.     Place    the    dihedrals    so    that    the 
plane  angles  coincide. 

2.  How  will  the  edges  lie  with  regard 
to  each  other?    A-uth. 

3.  How  will   the   respective   faces  lie 
with  regard  to  each  other  ?    Auth. 

Therefore — 

489.     COR.     //  two  dihedrals  are   equal   their  plane 
(ingles  are  equal. 


DIHEDRAL  ANGLES 


251 


PROPOSITION  XXIII. 

490.     THEOREM.     Two  dihedral  angles  have  the 
same  ratio  as  their  plane  angles. 


Given  two  dihedral  angles 
M  —  O—N  and  M'—O'—S', 
with  plane  angles  0  and  0'  re- 
spectively. 

LM-O-N       LO 


To  Prove 


LK'-O'-W     LO' 


Proof.     CASE  I.     When   the    plane    angles   are    com- 
mensurable. 

SUG.  1.  Divide  the  plane  angles  by  a  common 
unit  of  measure  and  suppose  the  unit  to  be  con- 
tained in  L  0  and  L  0'  m  and  n  times  respec- 
tively. What  then  is  the  ratio  of  L  0  to  L  0'  ? 

2.  Through  the  respective  edges  of  the 
two  dihedrals  and  the  various  division  lines  of 
the  two  plane  angles  pass  planes.     How  do  the 
dihedrals  thus  formed  compare  with  one  another  ? 
Auth.     How  many  of  them  are  there  in  L  M  — 
0  —  N1    In  L  M'  —0'  —  A7'? 

3.  What  is  the  ratio  of  the  two  dihe- 
drals?    Apply  §247.    a^y^ 

4.  Complete  the  demonstration. 

CASE  II.     When  the  plane  angles  are  incommensur- 
able. 

SUG.     Use  the  method  of  §  429  in  working  out 
a  demonstration. 
Therefore — 


252 


SOLID  GEOMETRY 


491.  MEASURE  OF  DIHEDRAL  ANGLES.    Since  dihedral 
angles  are  proportional  to  their  plane  angles,  they  are 
said  to  be  measured  by  their  plane  angles.    Thus  a  right 
dihedral  angle  is  a  dihedral  the  plane  angle  of  which  is 
a  right  angle.     Similarly  one  of  27°  is  one  the  plane 
angle  of  which  is  27°.     Dihedral  angles  are  adjacent, 
vertical,  acute,  obtuse,  etc.,  according  as  the  respective 
plane  angles  are  adjacent,  vertical,  acute,  obtuse,  etc. 

492.  PERPENDICULAR  PLANES.     Planes  which  form  a 
right  dihedral  angle  are  perpendicular  planes. 

PROPOSITION  XXIV. 

493.  THEOREM.     //  a  straight  line  is  perpen- 
dicular to  a  plane,  every  plane  containing  that 
line  is  perpendicular  to  the  plane. 


Given  line  EOL  plane  M  and  a  plane  N  containing 
EO  and  intersecting  M  in  OD. 
To  Prove  plane  N 1  plane  M. 

Proof.     Sue.  1.     What   must   be  known  to   make  N 
JiMf 

2.  In  plane  M  erect  OF  1  OD. 

3.  How    many    degrees    in     Z  EOF 'I 

4.  Z  EOF  is  the  plane  angle   of  the 
dihedral.     Why? 

5.  What   relation   does   plane   A7   bear 
to  plane  M<!    Why? 

Therefore — 


Why? 


DIHEDRAL  ANGLES  253 

1.  Vertical  dihedral  angles  are  equal. 

SUG.     Pass  a  plane  through  the  angles  perpendicular  to 
the  common  edges  and  compare  the  resulting  plane  angles. 

2.  Adjacent    dihedrals    are    supplementary    if    their   exterior 
faces  lie  in  the  same  plane. 

3.  Two   planes   which   bisect    respectively   two    vertical    dihe- 
drals form  one  and  the  same  plane. 

4.  A  plane  which  bisects   one   of  two  vertical  dihedrals   bi- 
sects the  other. 

PROPOSITION  XXV. 

494.  A  line  in  one  of  two  perpendicular  planes 
perpendicular  to  their  intersection  is  perpendicu- 
lar to  the  other. 


Given  plane   N  J_  plane  M  with  FD  the  line  of  inter- 
section and  in  plane  N  a  line  EG  _1_  FD. 
To  Prove  EG  1  plane  M. 

Proof.  SUG.  1.  What  must  be  proved  in  addition 
to  the  hypothesis  in  order  to  show  that  EG  is 
perpendicular  to  plane  M  ? 

2.  In  plane  M  draw  GH  LFD   at   G. 
Angle  EGH  is  the  plane  angle  of  the  dihedral. 
Why  ?    How  many  degrees  in  Z  EGH  ?    Why  ? 

3.  Complete     the     demonstration     by 
showing  that   EG  must  be  perpendicular  to  M. 
§448. 

Therefore — 


254 


SOLID  GEOMETRY 


495.  COR.  I.  //  two  planes  are 
perpendicular  to  each  other,  a  line 
perpendicular  to  one  of  them  at 
any  point  of  their  intersection  lies 
in  the  other. 


Given  M  and  N,  two  perpendicular  planes,  FD  their 
intersection,  and  OG  _]_ M  at  G,  a  point  in  FD. 
To  Prove  OG  lies  in  plane  X. 

Proof.  SUG.  1.  Suppose  OG  does  not  lie  in  N  and  that 
EG  is  the  line  in  A'  which  is  1.FD  at  0.  What 
relation  does  EG  bear  to  M  by  §  494? 

2.  How  many   perpendiculars  can  be 
erected  to  M  at  G  ? 

3.  AYhat  then  of  the  supposition? 
Therefore — 

496.  COR.  II.  //  two  planes  are  perpendicular,  a 
perpendicular  from  any  point  in  the  first  to  the  second 
lies  in  the  first. 

Proof.  Make  a  demonstration  following  the  plan  of 
§495. 

1.  If    a    plane   intersects    two    parallel    planes    the    alternate 
interior  dihedrals  are  equal. 

SUG.  A  plane  perpendicular  to  the  edge  EF  lies  how 
with  reference  to  edge  E'F'1 
Why?  It  cuts  the  parallel  planes 
M  and  M'  in  two  lines  as  a  and  «'. 
How  do  a  and  a'  lie  with  refer- 
ence to  each  other1?  Why?  Com- 
pare the  plane  angles  of  these 
dihedrals. 

2.  In  the  preceding  problem   prove  the  corresponding  dihe- 
drals equal  and  that  the  two   dihedrals  on  the  same  side  of  the 
secant  plane  are  supplementary. 


DIHEDRAL  ANGLES 
PROPOSITION  XXVI. 


255 


497.  THEOREM.  //  two  intersecting  planes  are 
each  perpendicular  to  a  third  plane,  their  inter- 
section is  perpendicular  to  the  third  plane. 


Given  two  planes  M  and  N,  each  1  plane  P  and  inter- 
secting in  the  line  OX. 
To  Prove  OX  1  plane  P. 

Proof.     Sue.  1.     At  point  0  which  is  common  to  all 
three  planes  erect  the  perpendicular  to  plane  P. 

2.  Where   does  this  perpendicular  lie 
with  reference  to  plane  M?   To  plane  JV?     §  495. 

3.  What  relation  does  this  perpendicu- 
lar then  bear  to  the  intersection  of  M  and  .V? 
Why? 

Therefore — 

1.  A    plane    which   is    perpendicular    to    the   intersection    of 
two  planes  is  perpendicular  to  the  planes. 

2.  The  plane  formed  by  a  line   and  its   projection   upon   a 
given  plane  is  perpendicular  to  the  given  plane. 

3.  The  projections   of   a  line   upon  two   parallel  planes   are 
equal. 

4.  What   condition   must   be   added   to    the    converse   of   the 
following  theorem  in  order  that  it  be  true?     If  two  parallel  planes 
are  cut  bv  a  third  plane  the  interior  angles  on  the  same  side  of 
the  secant  plane  are  supplementary.  See  ex,  1 1,  p.  268. 


256  SOLID  GEOMETRY 

t 

PROPOSITION  XXV II. 

498.  THEOREM.  Through  a  given  straight  line, 
oblique  or  parallel  to  a  given  plane,  one,  and  but 
one,  plane  can  be  passed  perpendicular  to  the 
given  plane. 

B/ 


Given  plane  M  with  line  a  oblique  or  parallel  to  M. 

I.  To  Prove  that  one  plane  can  be  passed  through  a 
and  1  M. 

Proof.     SUG.  1.     From  any  point  as  B  in  line  a  drop 
a  line  1  M,  as  BO. 

2.  BO    and    a    determine    a    plane,    N. 
Why? 

3.  What   relation   does  plane   N  bear 
to  plane  Mf    §  493. 

II.  To  Prove  that  N  is  the  only  plane  through  line 
a  perpendicular  to  M. 

SUG.  1.  Suppose  there  is  a  second  plane  P 
embracing  line  a  and  J_M.  What  relation  does 
line  a  bear  to  planes  N  and  Pf  §  439. 

2.  What  relation  must  line  a  then  bear 
to  plane  Mf    §  497. 

3.  Compare    this    conclusion    with    the 
hypothesis. 

Therefore — 


DIHEDRAL  ANGLES  257 

PROPOSITION  XXVIII. 

499.  THEOREM.  Every  point  in  a  plane  which 
bisects  a  dihedral  angle  is  equidistant  from  the 
faces  of  the  dihedral  angle. 


G 

Given  plane  P  bisecting  the  dihedral  formed  by  the 
planes  M  and  N,  X  any  point  in  plane  P,  and  XE  and 
XF  perpendiculars  from  X  to  M  and  N  respectively. 
To  Prove  XE  =  XF. 

Proof.  SUG.  1.  XE  and  XF  determine  a  plane. 
Let  0  represent  the  intersection  of  this  plane  with 
GH,  the  edge  of  the  dihedral.  Then  OE  and  OF 
will  be  the  intersections  of  this  plane  with  M  and 
N  respectively. 

2.  What  relation  does  plane  XEF  sus- 
tain to  plane  M  and  to  plane  JV?    To-  their  inter- 
section, GH?    §497. 

3.  XOE  and  XOF  are  the  plane  angles 
of   the    dihedrals    M  —  O  —  P    and    N  —  O  —  P 
respectively.     Why  ? 

4.  Compare      L  XOE     and      /.XOF, 
§  489;  A  XOE  and  A  XOF;  XE  and  XF. 

Therefore—- 
Query.    What   statement    in   the    theorem   makes    it 

necessary  that  XE    and  XF  be  perpendicular  to  M  and 

N  respectively? 


258  SOLID  GEOMETRY 

500.  COR.  I.    Every  point  equidistant  from  the  faces 
of  a  dihedral  is  in  the  bisector  of  the  angle. 

SUG.  With  the  figure  of  §  499,  assume 
XE  =  XF  and  prove  L  XOE  =  L  XOF.  Make 
a  full  detailed  demonstration. 

501.  COR.  II.    Every  point  not  in  the  plane  which 
bisects  a  dihedral  is  unequally  distant  from  the  faces  of 
the  dihedral. 

SUG.  Use  the  indirect  method  or  a  direct 
method  similar  to  that  of  ex.  3,  p.  37. 

PROPOSITION  XXIX. 

502.  THEOREM.     The    acute     angle    which    a 
straight  line  makes  with  its  projection  on  a  plane 
is  the  least  angle  the  line  makes  with  any  line  in 
the  plane  through  its  foot. 


Given  line  XB  meeting  plane  M  at  B,  BC  its  projec- 
tion on  M,  and  BD  any  line  in  M  through  B  other  than 
BC. 

To  Prove  L  XBC<  L  XBD. 

Proof.  SUG.  1.  From  any  point  in  XB,  as  A,  drop  a 
perpendicular,  AC,  to  M.  Where  will  (1  lip? 
Why? 

2.  On  the  second  line  through   />.  take 
BD  —  BC  and  join  A  and  D. 

3.  Compare  AC  with  AD.    Auth. 

4.  Compare      L  ABC     with      LAUD. 
§124. 

Therefore — 


'  DIHEDEAL  ANGLES  259 

1.  The  locus  of  points  in  space  equidistant  from  two  inter- 
secting planes  is  the  pair  of  planes  bisecting  the  dihedrals  formed 
by  the  given  planes. 

503.  ANGLE  OF  A  LINE  TO  A  PLANE.  The  acute  angle 
which  a  line  makes  with  its  projection  on  a  plane  is  the 
angle  of  the  line  to  the  plane. 

2.  The  supplement  of  the  angle  of  a  line  to  a  plane  is  the 
greatest  angle  which  the  line  makes  with  any  line  in  the  plane 
through  its  foot. 

3.  A  line  makes  equal  angles  with  parallel  planes. 

4.  There   is    one   and   but    one   perpendicular   which   can    be 
drawn  to  two  non  parallel  lines  which  are  not  in  the  same  plane. 

Given  two  lines,  b  and  c,  not  in  the  same  plane  and  not 
parallel.  Pass  a  line  d  through  one  of  them,  as  b,  parallel  to  the 
other,  c.  Pass  through  b  and  d  the  plane 
M.  How  does  M  lie  with  reference  to 
line  c?  Through  c  pass  a  plane  N  per- 
pendicular to  M.  Let  line  e  be  the  inter- 
section of  M  and  N.  Then  e  \\  c.  Why? 
Let  the  intersection  of  b  and  e  be  0  and 
at  0  erect  the  line  OPLM.  This  line  lies  in  N.  Why?  Hence 
OP  meets  c  and  OP  1  c.  Why? 

THEREFORE  there  is  at  least  one  such  common  perpendicular 
to  b  and  c. 

Suppose  there  is  another  such  perpendicular  as  EF.  The  plane 
of  line  c  and  EF  will  intersect  If  in  a  line  GF  and  be  1  to  M. 
Why?  Thus  there  will  be  two  planes  embracing  c  and  each  i  M, 
-\\uich  is  impossible?  Why? 

THEREFORE 

5.  In  the  preceding  theorem  the  common  perpendicular,  OP, 
is  the   shortest  line  which  can  be  drawn  between  the  two   given 
lines. 

6.  Find  a  point  X  equidistant  from  four  given  points  not 
in  the  same  plane. 

Sue.    Let  the  given  points  be  A,  B,  C,  D.    What  is  the 
locus  of  points  equidistant  from  A  and  B  ?     From  B  and  C? 


260  SOLID  GEOMETRY 

I 

From  A,  B,  and  C  ?     From  C  and  D®     Complete  the  demon- 
stration. 

504.  POLYHEDRAL    ANGLE.      Three    or    more    planes 
meeting  at  a  common  point  form  a  polyhedral  angle;  or 
simply  a  polyhedral. 

For  example,  A  — B(JD  represents  a  polyhedral  angle  formed  by 
three  planes.  The  common  point,  A,  is  the  vertex  of  the  angle, 
the  intersections  of  the  planes,  AB,  AC,  AD, 
are  the  edges  of  the  angle,  and  the  portions  of 
the  planes  included  between  the  successive 
edo-es*  are  the  faces  of  the  angle.  The  plane 
angles  formed  by  the  successive  edges  are  the 
face  angles  of  the  polyhedral.  The  face  angles 
and  the  dihedrals  of  the  successive  faces  are 
the  parts  of  the  polyhedral. 

505.  •  CONVEX  POLYHEDRAL  ANGLE.     If  the  intersec- 
tions of  a  plane  with  all  the  faces  of  a  polyhedral  form 
a  convex  polygon  the  polyhedral  is  a  convex  polyhedral 
angle. 

506.  CLASSIFICATION  OF  POLYHEDRAL  ANGLES.  A  poly- 
hedral with  three  faces  is  a  trihedral  angle,  or  a  trihe- 
dral; one  having  four  faces  is  a  tetrahedral  angle,  or 
tetrahedral;  etc. 

A  trihedral  angle  with  two  equal  face  angles  is  isosceles. 

507.  CONGRUENT  POLYHEDRAL   ANGLES.     Polyhedral 
angles  which  can  be  made  to  coincide  are  congruent  poly- 
hedral angles. 

For  two  polyhedrals  to  be  congruent,  it  is  necessary  that  the 
respective  parts  of  the  two  angles  be  equal  and  arranged  in  the 
same  order. 

508.  SYMMETRICAL  POLYHEDRAL  ANGLES.     Two  poly- 
hedral angles  having  the  dihedrals  ai^d  faces  angles  of 
the  one  equal  respectively  to  the  corresponding  parts  of 
the  other  but  arranged  in  the  reverse  order  are  symme- 
trical polyhedral  angles. 


POLYHEDRAL  ANGLES 

E' 


261 


In  general  two  symmetrical  polyhedral  angles  cannot 
be  made  to  coincide. 

Polyhedrals  E  —  FGH  and  E'  — F'G'H'  are  symmetrical  pro- 
vided L  PEG  =  L  F'E'G',  L  GEE  -  L  G'E'H',  L  HEF  =  L  H'E'F', 
dihedral  EF  —  dihedral  E'F',  etc.,  the  arrangements  of  the  re- 
spective parts  in  the  two  polyhedrals  being  opposite. 

As  an  illustration  of  the  symmetrical  relation  consider  a  pair  of 
gloves.  Two  gloves  for  the  same  hand  may  be  compared  to  equal 
polyhedrals  and  the  pair  to  symmetrical  polyhedrals. 

1.  Cut  from  pasteboard  two  figures  like  those  in  the  ac- 
companying figures,  creasing  them  on  the  dotted  lines.  Note 
that  all  parts  with  corresponding  notations  are  equal.  The  equal- 
ity of  the  "parts"  may  be  tested  by  superposition.  The  figures 
when  bent  into  polyhedrals  will  not  coincide. 
E 


2.  If     a     plane     be     passed 
through  either  diagonal  of  a  par- 
allelogram,   the    perpendiculars    to 
this  plane  from  the  extremities  of 
the  other  diagonal  are  equal. 

3.  Having  given  a  fixed  straight  line  and  two  points'  not  in 
the  line,  find  a  point  in  the  fixed  line  equally  distant  from  the 
given  points. 

4.  What  is  the  locus  of  a  point  equidistant  from  two  given 
parallel  planes  and  at  the  same  time  equidistant  from  two  given 
points  ? 


262  SOLID  GEOMETRY 

( 
PROPOSITION  XXX. 

509.     THEOREM.     The  sum  of  any  two  face  an- 
gles of  a  trihedral  angle  is  greater  than  the  third. 


Given  the  trihedral  A  —  BCD,  in  which  face  angle 
DAC  is  the  greatest. 

To  Prove  Z  CAB  +  Z  BAD  >  L  DAC. 

Proof.  SUG.  1.  It  is  unnecessary  to  prove  the  theo- 
rem for  the  cases  in  which  the  greatest  angle, 
LDACy  is  one  of  the  two  angles  added.  Why? 

2.  In  the  face  DAC  draw  a  line  AM 
equal  to  ABf  making   Z  MAD  equal  to   Z  DAB, 
and  through  B  and  M  pass  a  plane  cutting  the 
two  other  edges  in7  points  D  and  C. 

3.  Compare  A  BAD  with  A  MAD;  DM 
with.DB.    Auth. 

'  4.     Compare  DB  +  BC  and  BM  +  MC, 
compare  BC  and  MC.     Auth. 

5.  Compare      Z  BAC     with      Z  MAC. 
Auth. 

6.  Compare      /.  BAC  +  /.  BAD      with 
Z  DAC. 

Therefore— 

1.  What  is  the  locus  of  a  point  equidistant  from  two  given 
parallel  planes  and  at  the  same  time  equidistant  from  two  other 
parallel  planes? 


•POLYHEDRAL  ANGLES  263 

PROPOSITION   XXXI. 

510.  THEOREM.  The  sum  of  the  face  angles  of 
any  convex  polyhedral  angle  is  less  than  four 
right  angles. 


Given  a  polyhedral  angle  A  with  n  face  angles. 

To  Prove  the  sum  of  the  face  angles  about  vertex  A 
to  be  less  than  4  right  angles. 

Proof.  SUG.  1.  Pass  a  plane  through  the  polyhe- 
dral cutting  all  the  edges  in  the  points  B,  C,  D,  F, 
etc.  This  cross  section  polygon  will  have  n  sides 
and  the  plane  will  form  with  the  faces  n  "face" 
triangles  with  a  common  vertex  A. 

2.  Let  0  be  any  point  within  this  poly- 
gon and  join  0  to  the  n  vertices  of  the  polygon, 
forming  n  "base"  triangles  with  a  common  ver- 
tex 0. 

3.  The  sum  of  the  angles  of  the  face 
triangles   equals  the   sum  of  the   angles  of  the 
base  triangles.     Why? 

4.  Z  ABC  +  L  ABF    >     L  CBF, 
L  ACS  +  L  ACD  >  L  BCD,  etc. 

5.  Compare   then  the   sum   of  all  the 
base  angles  of  the  face  A  with  the  sum  of  all  the 
base  angles  of  the  base  A. 

6.  Compare  the  sum  of  the  face  angles 
at  A  with  the  sum  of  the  angles  about  0. 


264 


SOLID  GEOMETRY 


7.     Compare  the  sum  of  the  angles  at 
A  with  four  rt.  A. 
Therefore — 

PROPOSITION   XXXII. 

511.  THEOREM.  //  two  trihedral  angles  have 
the  three  face  angles  of  the  one  equal  respectively 
to  the  three  face  angles  of  the  other,  the  corre- 
sponding dihedral  angles  are  equal. 


Given   two    trihedrals,   A    and   A',   with 
Z  BAG  =  L  B'A'C',   L  CAD  =  L  C'A'D', 
L  DAB  =  Z  D'A'B'. 

To  Prove   dihedral    AB  =  dhl   A'B', 
dhl  AC  =  dhl  A'C',  and  dhl  AD  =  dhlA'D'. 

Proof.  SUG.  1.  Take  points  on  the  six  edges  so 
that  AB  —  AC  =  AD  =  A'B'  =  A'C'  =  A'D'  and 
AM  =  A'M'.  Through  B,  C,  D  and  B',  C',  D' 
pass  planes.  In  the  faces  BAC  and  BAD  re- 
spectively draw  MN  and  MP  each  l  AB.  MP  and 
MN  can  meet  BC  and  £Z>  in  N  and  P  respectively 
for  A.  AB  C  and  AB  Dare  acute.  Why?  Similarly 
draw  lines  M'N'  and  M'P'  through  point  M'. 

2.  The    dihedrals    AB    and   4' 5'    are 
equal  if  Z  PM#=  Z  P'M'N'.    Why? 

3.  Compare    A  AB(7   with    AA'B'C',- 
L  ABC  with  Z  A'B'C',  BC  with  B'C'.    Auth. 


POLYHEDRAL  ANGLES  265 

4.  Draw   the  similar  conclusions  from 
&ABD   and  A'B'D'.     Also  from   A  ACD   and 
A'C'D'. 

5.  Compare    A  BMP    with    kB'M'P', 
BP  with  B'P' ;  MP  with  J/'P'.       Auth.      Draw 
similar  conclusions  from  A  BMN  and  A  B'M'N'. 

6.  Compare    A  BCD    with    AB'C'D'; 
Z  C5D  with  Z  C'B'D';  A  £PA7  with   A  tf'P'tf'; 
AT  with  N'P'.     Auth 's. 

7.  Compare    &  XMP    with    A  N'M'P', 
Z  J/  with  Z  jl/'.  Auth. 

8.  Compare  dihedral  AB  with  dihedral 
A'£'.     Auth. 

9.  By  similar   arguments  compare   di- 
hedrals AC  and  A'C",  AD  and  A'ZT. 

Therefore— 

PROPOSITION   XXXIII. 

512.  THEOREM.  //  two  trihedral  angles  have 
the  three  face  angles  of  the  one  equal  respectively 
to  the  three  face  angles  of  the  other,  they  are 
cither  congruent  or  symmetrical. 


Given  two  trihedrals,  0  and  0',  with  ZJfOP—  Z.¥'0'P', 
ZPON  =  Z.P'0'N',  and  /.NOM  =  Z.V'OMT. 

I.  To  Prove  trihedrals  0  and  0'  congruent,  the  an- 
gles being  arranged  in  the  same  order. 


266  SOLID  GEOMETRY 

Proof.  SUG.  1.  Place  0  upon  0'  so  that  /.MOP 
coincides  with  LWO'P'  and  the  two  edges  ON 
and  O'N'  are  on  the  same  side  of  the  plane  MOP. 

2.  How  does  plane  PON  lie  with  ref- 
erence to  plane  P'O'N't    Plane  NOM  with  ref- 
erence to  plane  N'O'M'f    §511. 

3.  Where  does  line  ON  lie  with  respect 
to  O'N"!    Why? 

II.  To  Prove  trihedrals  0  and  0'  symmetrical,  the 
angles  being  arranged  in  reverse  order. 

Proof.  By  §  511  the  corresponding  dihedrals  are 
equal.  The  parts  are  therefore  respectively  equal  and 
by  hypothesis  the  order  of  the  angles  is  different  in 
0  and  0'. 

Therefore— §  508. 

If  a  line  and  a  plane  are  parallel,  a  line  drawn  from  any  point 
in  the  plane  parallel  to  the  line  will  lie  in  the  plane. 

1.  Parallel  lines  intersecting  a  plane  make  equal  angles  with 
the  plane. 

2.  What   is  the   locus   in    space    of   points   equidistant   from 
the  sides  of  a  plane  angle? 

3.  Bisect  a  dihedral  angle. 

4.  What  is  the  locus  of  a  point  in  a  plane  equidistant  from 
two  points  without  the  plane? 

5.  Find  a  point  X  in  a  given  line  and  equidistant  from  two 
points  without  the  line. 

6.  Find  a   point   X  in   a  plane  and  equidistant   from   three 
points  without  the  plane. 

7.  What  is  the  locus  of  a  point  X  in  a  given  plane,  a  given 
distance  from  a  point  not  in  that  plane? 

8.  Find  a  point  X  in  a  plane  and  equidistant  from  the  three 
edges  of  a  trihedral. 


•POLYHEDRAL  ANGLES 


267 


PROPOSITION  XIV. 

513.     THEOREM.     Two     symmetrical     isosceles 
trihedral  angles  are  congruent. 


Given  two  symmetrical  isosceles  trihedrals  0  and  0' 
in  which  L  MOP  =  L  M'O'P',  L  PON  =  L  P'O'N', 
L  NOM  =  L  N'O'M',  etc.,  and  L  NOM=/L  PON  and 
N'O'M'  =  L  P'O'N1 '. 

To  Prove  trihedrals  0  and  0'  congruent. 

Proof.  SUG.  1.  On  account  of  the  symmetry,  which 
face  angles  and  which  dihedrals  are  equal? 

2.  Because  each  is  isosceles,  which  face 
angles  are  equal? 

3.  Compare     Z  NOM    with     Z  P'O'N'; 
Z  PON  with  Z  N'O'M'. 

4.  From   step    3,    rearrange   the   equal 
parts  so  that  the  order  is  the  same  for  0  and  0 ' . 
§507. 

Therefore— 

514.  VERTICAL  POLYHEDRAL  ANGLES.  When  two 
polyhedrals  are  so  placed  that  they  have  a  common  ver- 
tex and  the  sides  of  the  one  are  extensions  through  the 
vertex  of  the  sides  of  the  other,  they  are  vertical  poly- 
hedral angles. 

1.  Vertical  trihedrals   are  symmetrical. 

2.  Vertical    polyhedrals    are    symmetrical. 

3.  What  is  the   locus  of  points  equidistant   from   the   faces 
of  a  trihedral?     §499. 


268  SOLID  GEOMETRY 

4.  What  is  the  greatest  number  of  equilateral  triangles  which 
can  be  put  together  to  form  a  convex  polyhedral  angle  I 

5.  How  many  different  sized  polyhedrals  can  be  formed  from 
equilateral  triangles  ? 

6.  How    many    different    sized    polyhedrals    can    be    formed 
from    squares?      From   .regular    pentagons?      From    regular    hexa- 
gons?    §  509. 

7.  How  many  different  polyhedrals  can  be  constructed  from 
regular  polygons? 

8.  Can  a  mosaic  or  patch  work  pattern  be  constructed  from 
regular    triangles?      From    squares?      From    regular    pentagons? 
From  regular  hexagons?     From  regular  heptagons?     From   regu- 
lar n-gons?     Give  reasons  for  each  conclusion. 

9.  The  foundation  which  bee  keepers  supply  for  the  bees  to 
build   the  honey   comb   upon   is   a   mosaic   constructed   of   regular 
polygons  as  near  as  possible  to  the  shape  of  a  circle.     Of  what 
polygons  is  it  formed? 

10.  A  circle  would  be  a  better  shaped  base  for  the  body  of 
the  bee.     Will  the  circle  or  polygon  form  of  construction  require 
the  less  material? 

It  is  assumed  in  the  preceding  exercises  that  the  amount  of 
material  used  in  making  the  cells  is  proportional  to  the  amount 
used  in  making  the  base. 

11.  Are   two   planes  perpendicular   to   the   same   plane   neces- 
sarily parallel?     Are  two   lines  perpendicular  to  the   same   plane 
parallel?     Are  two  planes  perpendicular  to  the  same  line  parallel? 

12.  Which  is  the  longer,  an  oblique  sect  or  its  projection  on 
a  plane? 

13.  If  three  non  parallel  planes  are  each   perpendicular   to  a 
fourth  plane  their  three  lines  of  intersection  are  parallel. 

14.  If  two  parallel  planes  intersect  a  dihedral,  the  respective 
lines  of  intersection  form  equal  angles. 

15.  If  the  mid  points  of  the  adjacent  sides  of  a  skew  quad- 
rilateral (i.  e.  one    the  four  vertices    of  which  are  not  in  the  same 
plane)    are*  joined    by    straight    lines,    the    figure    enclosed    is    a 
parallelogram. 


EXERCISES  UGi) 

515.  TKIRECTANGULAR  TRIHEDRAL.  A  trihedral  angle 
all  of  the  face  angles  of  which  are  right  angles  is  a  tri- 
rectangular  trihedral  angle. 

16.  In   a   trirectangular   trihedral  the   dihedrals   opposite   the 
equal  face  angles  are  equal.     Why?     Is  this  true  of  any  isosceles 
trihedral? 

REVIEW 

1 7.  State  in  a  theorem  a  possible  condition  by  which  one  line 
can  be  proved  parallel  to  another;    a  line  can  be  proved  parallel 
to  a  plane;  a  plane  be  proved  parallel  to  a  second  plane. 

18.  State  in  a  theorem  a  condition  by  which  a  line  may  be 
proved  perpendicular  to  a  second  line;   to  a  plane;   a  plane   can 
be  proved  perpendicular  to  a  second  plane. 

19.  Stp.te  in  a  theorem   a  condition  by  which  two  trihedrals 
can  be  proved  congruent ;    can  be  proved  symmetrical ;   two  sym- 
metrical  trihedrals  can  be  proved  congruent. 

20.  What   is  the   locus    of   points  equidistant   from   the   ver- 
tices of  a  triangle? 

21.  Locate    a   point   in    a    given    plane    equidistant    from    all 
points  of  a  circle  which  is  in  another  plane. 

22.  Find  a  point  X  which  is   equidistant   from  two  parallel 
planes,  from  two  intersecting  planes,  and  from  two  given  points. 

23.  If  two  supplementary  adjacent  dihedrals  are  bisected  by 
planes,  the  bisecting  planes  form  a  right  dihedral. 

24.  If  from  any  point  within  a  dihedral  angle  perpendiculars 
are  drawn  to  the  faces  of  the  dihedral,  the  angle  formed  is   the 
supplement  of  the  dihedral. 

25.  If  from  any  point  without  a  dihedral  perpendiculars  are 
drawn  to  the  faces  of  the  dihedral,  the  angle  formed  by  the  per- 
pendiculars is  equal  to  the   dihedral. 

26.  Two  trihedrals  having  two  face  angles  and  the  included 
dihedral  of  the  one  equal  respectively  to  the  corresponding  parts 
of  the  other  are  either  congruent  or  symmetrical. 

27.  Two    trihedrals    having    two    dihedrals    and    the    included 
face    angle   of  the   one   equal   to    the   corresponding   p'arts   of    the 
other  are  either  congruent   or  symmetrical. 


CHAPTER  VIII, 


POLYHEDRONS. 

516.  POLYHEDRON.  A  geometric  solid  bounded  by 
planes  is  a.  polyhedron.  The  intersections  of  the  planes 
are  the  edges;  the  intersections  of  the  edges  are  the  ver- 
tices; the  portions  of  the  planes  bounded  by  the  edges 
are  the  faces;  the  face  upon  which  the  polyhedron  is 
supposed  to  rest  is  its  base.  Any  face  may  be  taken  as 
the  base.  Any  straight  line  connecting  two  vertices  not 
in  the  same  facets  a  diagonal  of  the  polyhedron. 


517.  POLYHEDRONS     CLASSIFIED.       Polyhedrons     are 
classified  according  to  the  number  of  faces.    One  of  four 
faces  is  a  tetrahedron,  of  six  faces  is  a  hexahedron,  of 
eight  faces  is  an  octohedron,  of  twelve  faces  is  a  dodec- 
ahedron, of  twenty  faces  is  an  icosahedron,  etc. 

1.        What  is  the  least  number  of  faces  possible  for  a  poly- 
hedron! 

518.  £LANE  SECTION  OP  A  POLYHEDRON.     The  inter- 
section of  a  plane  and  a  polyhedron  is  a  plane  section  or 
a  section  of  Mio  polyhedron.     FE   h  ;i   plmio  section  of 
polyhedron. 


POLYHEDRONS  271 

519.  CONVEX  POLYHEDRON.     A  polyhedron  such  that 
every  plane  see  lion  is  a  convex  polygon  is  a  convex  poly- 
hedron. 

Only  convex  polyhedrons  will  be  considered  in  this  book. 

520.  PRISM.     A  polyhedron  bounded  by  two  paral- 
lel planes  and  a  group  of  planes  the  intersections  of  which 

•are  parallel  lines  is  a  prism. 

rP 


The  faces  formed  by  the  parallel  planes  are  the  bases, 
usually  designated  as  upper  and  lower  base.  The  other 
faces  are  the  lateral  faces.  The  intersections  of  the 
lateral  faces  with  each  other  are  the  lateral  edges  and 
their  intersections  with  the  bases  are  the  basal  edges. 

521.  RIGHT   SECTION.     A  section  of  a  prism  by  a 
plane  perpendicular  to  the  lateral  edges  is  a  right  sec- 
tion. 

522.  OBLIQUE  SECTION.     A  section  of  a  prism  by  a 
plane  which  is  oblique  to  the  lateral  edges  is  an  oblique 
section. 

523.  ALTITUDE  OF  A  PRISM.     The  distance  between 
the  bases  of  a  prism  is  its  altitude.    §  477. 

524.  CLASSIFICATION  OF  PRISMS.    Prisms  are  classified 
as  triangular,  quadrangular,  etc.,  according  as  their  bases 
are  triangles,  quadrilaterals,  etc. 

PRELIMINARY  THEOREMS 

525.  THEOREM  I.     The  lateral  edges  of  a  prism  are 
equal.     §   479. 

526.  THEOREM  IT.     The  lateral  faces  of  a  prism  are 
parallelograms.     §  520. 


272  SOLID  GEOMETRY 

527.  THEOREM  III.     The  acute  angles  made  by  the 
lateral  edges  with  the  planes  of  the  bases  are  equal.  §  503. 

528.  RIGHT  PRISM.    A  prism  in  which  the 
lateral  edges  are  perpendicular  to  the  bases  is 
a  right  prism. 

The    edge    of    a    right    prism    is    also    its 
altitude. 

529.  OBLIQUE  PRISM.     A  prism  in  which  the  lateral 
edges  are  not  perpendicular  to  the  bases  is  an  oblique 
prism. 

530.  REGULAR  PRISM.     A  right  prism  the  bases  of 
which  are  regular  polygons  is  a  regular  prism. 

PKOPOSITION   I. 

531.  THEOKEM.     Sections  of  a  prism  made  by 
parallel  planes  are  congruent  polygons. 


Given  ABC  and  A'B'C',  two  parallel  sections  of  a 
prism  P. 

To  Prove  ABC  =  A'B'C'. 

Proof.  SUG.  1.  What  two  relations  do  the  lines  AB, 
BC,  CD,....  bear  to  the  lines  A'B',  B'C',  C'D', 
....  respectively  ?  §  478. 

2.     Compare    A  ABC,   BCD,   etc.,    with 
A  A'B'C',  B'C'D',  etc.,  respectively.     Auth. 
Therefore — 


PRISMS 


273 


532.  COR.   I.     The   bases  of  a  prism  are  congruent 
polygons. 

533.  COR.  II.    A  section  of  a  prism  made  by  a  plane 
parallel  to  th<    bas<    is  congruent  to  the  base. 

534.  COR.   III.     All   right  sections  of  a  prism  are 
congruent  polygons. 

PROPOSITION  II. 

535.  THEOREM.     The  lateral  area  of  a  prism  is 
equal  to  the  product  of  the  perimeter  of  a  right 
section  and  a  lateral  edge. 


Given  the  prism  M  with  p  the  perimeter  and  at,  a2>  av 
etc.,  the  successive  sides  of  a  right  section,  e  the  length 
of  the  equal  lateral  edges,  and  S  the  lateral  area. 
To  Prove  S  =  e  x  p, 

Proof.  SUG.  1.  How  do  the  successive  sides  of  the 
right  section  lie  with  respect  to  the  successive 
edges  which  they  intersect  ? 

2.  What  is  the  area  of  the  face  EF  in 
terms  of  at  and  e  ? 

3.  Express    the    area    of    each    lateral 
face. 

•4.     By  adding  these  areas  find  the  total 
lateral  area,  8,  in  terms  of  a^  a.,,  «3,  etc.,  and  e 
and  then  in  terms  of  e  and  p. 
Therefore — 


274  SOLID  GEOMETRY 

536.  COR.     The  lateral  area  of  a  right  prism  equals 
the  product  of  tin   perimeter  of  the  base  and  a  lateral 
edge. 

PROPOSITION  III. 

537.  THEOREM.     //  two  prisms  hare  the  three 
faces  of  a  trihedral  of  one  congruent  to  the  three 
faces  of  a   trihedral  of  the  other  and  similarly 
placed,  the  prisms  are  congruent. 


E        B  E'        D' 

Given  two  prisms  P  and  P'  in  which  the  three  faces 
AB,  AC,  AD  forming  the  trihedral  A  are  respectively 
congruent  to  the  three  faces  A'B' ;  A'C' ,  A' D'  forming 
the  trihedral  A'  and  are  similarly  placed. 

To  Prove  PEP'. 

Proof.     SUG.  1.     Compare  trihedrals  A  and  A',  §512. 

2.  Apply    P'  to  P    so  that  face  A'B' 
coincides  with  face  AB.    Why  can^  this  be  done  ? 

3.  In  what  plane  does  face  A'C'  fall? 
FaceA'D'?     Why?     §511. 

4.  Where  does  line  A' E'  fall?    Why.' 
Where  do  points  E',C',D'  fall?    Why?    Where 
does  plane  C'D'  fall?    Why? 

5.  Compare  faces  C'D'  and  CD.    §  532. 

6.  Complete  the  superposition.     §  520. 
Therefore— 


PRISMS 


538.  COR.    I.      Two    right   prisms    arc    congruent    if 
their  altitudes  arc  equal  and  their  bases  congruent. 

539.  TRUNCATED  PRISM.     The  portion  of  a  prism  in- 
cluded between  the  base  and  a  section  made  by  a  plane 
not  parallel  to  the  base  is  a  truncated  prism. 

540.  COR.  II.     //    two    truncated   prisms    have    the 
flu'cc  faces  about  a  trihedral  of  one  congruent  respect- 
ively to  the  three  faces  about  a  trihedral  of  the  other  and 
similarly  placed,  the  truncated  prisms  are  congruent. 

Proof.     Draw  figures  and  make  a  proof  according  to 
the  method  of  §  537. 

PROPOSITION  IV. 

541.  THEOREM.     An  oblique  prism  is  equal  to 
a  right  pri.siu  the  base  of  which  /*  a  right  section 
of  the  oblique  prism  and  the  altitude  of  which  is 
equal  to  the  edge  of  the  oblique  prism. 


\ 


"M L— 'D' 

Given  an  oblique  prism,  BC,  with  a  right  section  A'B' 
and  a  lateral  edge  AC. 

To  Prove  prism  BC  equal  to  a  right  prism  with  base 
A'B'  and  an  altitude  equal  to  AC. 

Proof.  SUG.  1.  Extend  the  lateral  edges  making 
A'C'  =AC,  and  through  C'  pass  a  plane  I!  plane 
A'B' .  Then  A' D'  is  a  right  prism. 


276  SOLID  GEOMETRY 

2.  Compare  the  faces  about  the  trihe- 
dral A  with  the  faces  about  the  trihedral  C. 

3.  Compare  the  truncated  prism  ABJ 
with  the  truncated  prism  CD ' . 

4.  Compare     prism     BC    with    prism 
B'C'. 

Therefore— 

542.  PARALLELOPIPEDS.  A  prism  the  bases 
of  which  are  parallelograms  is  a  parallelo- 
piped. 


543.  RIGHT  PARALLELOPIPED.     A  parallelepiped  the 
edges  of  which  are  perpendicular  to  the  bases  is  a  right 
parallelepiped. 

544.  RECTANGULAR  PARALLELOPIPED.     A  right  paral- 
lelopiped     the  bases  of  which  are  rectangles  is  a  rec- 
tangular parallelepiped. 


545.  CUBE.  A  rectangular  parallele- 
piped all  the  faces  of  which  are  squares  is 
a  cube. 


546.  VOLUME  OF  A  POLYHEDRON.     The  measure  of  a 
polyhedron  in  terms  of  some  other  polyhedron  taken  as 
the  unit  of  measure  is  the  volume  of  the  polyhedron. 

547.  UNIT  OF  MEASURE  FOR  VOLUME.     A  cube  with 
an  edge  equal  to  a  given  linear  unit  is  the  unit  of  meas- 
ure for  volume. 

If   a   polyhedron   contains   a   cubic   inch   twenty-five   times,   the 
volume  of  the  polyhedron  is  twenty-five  cubic  inches. 

548.     PRELIMINARY  THEOREMS. 

THEOREM  T.     All  faces  of  a  parallelepiped  are  paral- 
lelograms. 


PRISMS  277 

THEOREM  II.  All  faces  of  a  rectangular  parallelo- 
piped  are  rectangles. 

THEOREM  III.  All  faces  of  a  cube  are  congruent 
squares. 

THEOREM  IV.  Any  pair  of  opposite  faces  of  a  paral- 
lelopiped  man  be  taken  as  bases. 

PROPOSITION   V. 

549.  THEOLEM.  Opposite  faces  of  a  parallelo- 
piped  are  parallel  and  congruent  parallelograms, 
(ittf/  any  section  by  a  plane  cutting  four  parallel 
edges  is  a  parallelogram. 


Given  a  parallelepiped  AG  with  bases  AC  and  EG, 
AH  and  EG  being  one  pair  of  opposite  faces,  and  FHA 
a  section  made  by  a  plane  cutting  four  parallel  edges. 

To  Prove  faces  AH  and  BG  parallel  and  congruent 
parallelograms,  and  FD  a  CU. 

Proof.     SUG.  1.  AH    and    BG    are    parallelograms. 
Why? 

2.  Prove  them  congruent. 

3.  Prove  them  parallel.     §  478. 

4.  Prove  opposite  sides  of  FD  parallel. 
Therefore— 

1.  Every  section  of  a  prism  made  by  a  plane  parallel  to  a 
lateral  edge  is  a  parallelogram. 


278  SOLID  GEOMETRY 

1.  If  from  any  point  iii  space  perpendiculars  are  drawn  to 
the   lateral   faces   of  a   prism,   or   to    the   lateral   faces   extended, 
these  perpendiculars  are  all  in  the  same  plane. 

2.  Any    straight    line    drawn    through   the    middle    point    of 
any    diagonal    of    a   parallelopiped,    terminating    in    two    opposite 
faces  is  bisected  at  that  point. 

3.  The   four   diagonals   of   a   rectangular   parallelopiped    aro 
equal  to  one  another. 

4.  The  square  of  a  diagonal  of  a  rectangular  ptxallelopiped 
is  equal  to  the  sum  of  the  squares  on  three  concurrent  edges. 

5.  The  sum  of  the  squares  upon  the  four  diagonals  of  a  rect- 
angular parallelopiped   is  equal  to  the  sum   of  the   squares  upon 
the  twelve  edges. 

6.  Prove    the    preceding    exercise     for    any     parallelopiped. 
S8  337,  338. 

7.  Prove  that  the  lateral  area  of  a  right  prism  is  less  than 
the  lateral  area  of  any  oblique  prism  having  the  same  base  and 
an  equal  altitude. 

Sue.  Draw  an  oblique  and  a  right  prism  upon  the  game 
base  with  equal  altitudes.  No  face  of  the  oblique  prism 
has  a  less  altitude  than  the  corresponding  face  of  the 
right  prism.  Why  ?  Some  faces  of  the  oblique  prism 
must  have  altitudes  greater  than  those  of  the  correspond- 
ing faces  of  the  right  prism.  Why?  The  altitudes  of  the 
faces  are  defined  with  respect  to  the  common  bases. 

8.  Make   a  list   of  theorems   on   polyhedrals   and  in   connec- 
tion with  each  write  out  a  plane  geometry  theorem  which  closely 
corresponds  to  it. 

9.  From  two  points  on  the  same  side  of  a  plane  draw  two 
lines  to  a  point  in  the  plane  that  shall  make   equal   angles   with 
the   plane. 

10.  The  lines  joining  the  mid  points  of  opposite  sides  of  a 
skew  quadrilateral  bisect  each  other.     Ex. 

11.  A  plane  perpendicular  to  a  line  in  a  plane  is  perpendicu 
lar  to  that  plane. 

12.  If  two  or  more  planes  intersect  in  one  straight  line,  the 
perpendiculars  drawn  to  them  from  any  one  point  lie  in  the  same 
plane. 


PRISMS  279 

PROPOSITION   VI. 

550.  THEOREM.  A  plane  passed  thro  null  tico 
diagonally  opposite  edges  of  a  parallelepiped  di- 
vides the  parallelepiped  Into  two  equal  triangular 
prisms. 


D'  B 

Given  parallelepipeds  CD'  divided  by  a  plane  through 
two  opposite  edges   CC'   and  DD'   into  two  triangular 
prisms  A'B'C'- A  and  B'C'D'-B. 
To  Prove  A'B'C'- A  =  B'C'D'- B. 
Proof.     SUG.  1.     A  right  section  as  HF  intersects  a 
CH  on  the  parallelepiped  and  this  is  divided  into 
two  congruent  A  by  the  plane  CD'.     Why? 

2.  Compare    A-prism   A'C'D'—  A  with 
a  right  prism  on  EHF  as  base  with  an  altitude 
equal    to    D'D.      Compare    A-prism    B'C'D'—  B 
with  a  right  prism  on  GHF  as  base  with  an  alti- 
tude equal  to  C'C.    Auth. 

3.  Compare    these    two    right    prisms. 
§538. 

4.  Compare  A'C'D' -A  with  B'C'D'-B. 
Therefore — 

In  the  above  figure  which  prisms  are  congruent  and  which  are 
only  equal? 

1.  In  550  prove  A'B'C'  — A  and  B'C'D'  —  B    symmetrical. 

2.  The     four     diagonals     of     a     parallelepiped    bisect     each 
other. 


280 


SOLID  GEOMETRY 


PROPOSITION    VII. 

551.  THEOREM.  Two  rectangular  parallelo- 
pipeds  having  congruent  bases  are  proportional 
to  their  altitudes. 


Given  two  rectangular  parallelepipeds  P  and  P'  with 
congruent  bases  and  altitudes  a  and  a'  respectively. 

P      a 

To  Prove  —  =  — . 
P'     a 

Proof.     Case  I.     a  and  a'  commensurable. 

SUG.  1.  Divide  the  altitudes  by  a  common 
unit  of  measure,  supposing  it  to  be  contained  m 
and  m'  times  in  a  and  a'  respectively.  What  is 
the  ratio  of  the  altitudes? 

2.  Through  the  points  of  division  of  the 
altitudes  pass  planes  parallel  to  the  bases.    What 
kind   of  parallelepipeds  are   formed?     Compare 
them  in  number  and  volume.    Auth.    What  is  the 
ratio  of  P  and  P' ? 

3.  Compare  the   ratios  of  the   altitudes 
and  the  parallelepipeds. 

Case  II.    a  and  a'  incommensurable. 

SUG.  1.  Divide  the  altitudes  a  and  a'  by  any 
unit  commensurable  with  a'.  In  the  division  of 
a  there  will  be  a  remainder  x  less  than  the  unit. 
Why?  By  taking  this  unit  smaller  and  smaller 


PRISMS 


281 


this  remainder  may  be  made  to  decrease  indefi- 
nitely.   Why  ? 


M1 


2.  Through  this  last  point  of  division,  M, 
pass  a  plane  parallel  to  the  base.    As  the  unit  in 
use  is  decreased  what  change  takes  place  in  the 
altitude  031?     What  change  takes  place  in  the 
parallelepiped  MN1     In  the  parallelepiped  with 
altitude  xt     What    is  the    limit    of    the    varia- 
ble   altitude    OMf      Of    the    variable    parallele- 
piped MA7? 

0.17  .a         .  MN  .  P 

3.  ••--  =  —  and  - 

a'       a  P'       P' 

MX      OM     Wl 

4.  — == .    Why? 

P'        a 

5.  •'•   —  =  -.    §426. 

P'     a 

Therefore— 

552.  DIMENSIONS  OF  A  PARALLELOPIPED.  The  dimen- 
sions of  a  parallelepiped  are  its  altitude  and  the  base 
c^nd  altitude  of  its  base.  Consequently  the  three  dimen- 
sions of  a  rectangular  parallelepiped  are  any  three  con- 
current edges. 


282 


SOLID  GEOMETRY 
PROPOSITION   VIII. 


553.  THEOREM.  Two  rectangular  parallelepi- 
peds having  equal  altitudes  are  proportional  to 
iheir  bases. 


\ 


P' 


Given  two  rectangular  parallelepipeds  P  and  P'  with 
equal  altitudes  a,  their  bases  B  and  B'  having  the  di- 
mensions b,  c  and  b' ,  c'  respectively. 

P       B 
To  Prove   —  =  —  • 

Proof.  SUG.  1.  Construct  a  third  rectangular  par- 
allelepiped N  with  the  altitude  a  and  base  with 
dimensions  b' ,c. 


2. 


P      b         ,  N      c 

-  =  —  and  —  =  — 


N      b' 


P' 


.,,., 

V,  hy  ? 


3.     Hence 


-  =     c-  =    -.     Why? 
P'      ])'c       B' 


Therefore— 


1.  A  gable  for  a  dormer  window  is  triangular  in  shape  with 
an  angle  of  60°  at  the  ridge.     If  the  length  of  the  rafters  is  6  ft., 
what  is  the  area  of  the  largest  circular  window  that  can  be  in- 
serted, making  allowance  for  a  four  inch  frame  around  the  win- 
dow! 

2.  The  diagonals  of  a  rectangular  parallelepiped  are  equal. 

3.  Determine  a  point  in  one  side  of  a  triangle  equally 
from  the  other  two  sides. 


PRISMS 


283 


PROPOSITION  IX. 

554.  THEOREM.  Two  rectangular  parallelo- 
pipeds  are  to  each  other  as  the  product  of  their 
three  dimensions. 


I 


Given  two  rectangular  parallelepipeds  P  and  P',  with 
dimensions  a,  b,  c  and  a',  b'  ,  c'  respectively. 

P       axbxc 
To  Prove  —  —  — 

P'     a'xb'xc' 

Proof.     SUG.  1.     Let    N  be   a  rectangular  parallele- 
piped with  the  dimensions  a,  b,  c'  . 

j  axb     „,. 

2.     Then    —  =  -  and  —  =  -    —  •  W  hy  ? 


PC       jN        axb 

—  =  -  and  —  =  -    —  • 
N     c'         P'      a'xb' 


3.      -=-.      Why? 

P'      a'xb'xc' 
Therefore  — 

555.     COR.  I.     The  volume  of  a  rectangular  parallele- 
piped is  the  product  of  its  three  dimension*. 

Given  a  rectangular  parallelepiped  P  with  dimensions 
a,  6,  c. 

To  Prove  Vol.  P  =  abc. 

Proof.  SUG.  1.  Take  as  the  unit  of  volume  a  cube  U 
with  an  edge  equal  to  the  linear  unit  in  which  a, 
6,  c  are  expressed. 


284  SOLID  GEOMETRY 

i 

P      axbxc 

2.  -  =  abc. 
U     Ixlxl 

3.  •'•  Vol.  P  =  abc.  U. 
Therefore— 

In  the  applications  of  this  theorem  the  three  dimensions  must 
be  expressed  in  terms  of  the  same  linear  unit  and  the  unit  of 
volume  must  be  a  cube  with  an  edge  equal  to  this  linear  unit. 

556.  COR.     Two   rectangular  parallelepipeds  having 
two  dimensions  respectively  equal  are  to  each  other  as 
their  third  dimensions. 

557.  COR.  II.     The  volume  of  a  cube  equals  the  cube 
of  its  edge. 

558.  By  comparison  of  the  theorem  and  the  defini- 
tion of  volume  (§546)  it  will  be  observed  that  the  vol- 
ume  of   a   rectangular   parallelepiped   is   equal    to   the 
product  of  the  measures  of  the  three  edges  meeting  at 
any  vertex  times  the  unit  of  volume.     The  expression 
' '  product  of  the  three  dimensions ' '  is    un- 
derstood to  mean  the  product  of  the  mea- 
sures of  these  lines.  §  361  note. 

When  each  dimension  of  the  rectangu- 
lar parallelepiped  is  divisible  by  the  linear 
unit  which  is  the  edge  of  the  unit  vol- 


ume, the  truth  of  the  theorem  on  volume  may  be  shown 
by  dividing  the  parallelepiped  into  unit  cubes. 

1.  What  is  the  volume  of  a  cube  the  edge  of  which  is  7  in.? 

2.  What   is   the   volume   of  a  rectangular  parallelepiped  the 
dimensions  of  which  are  4  in.,  7  in.,  12  in.?      6  ft.  VI 2  ft.,  \/TS"  ft.? 
VlT,  \'T8,  V24?      V87V24,  vT2~? 

3.  The  edges  of  a  rectangular  parallelopiped  are  8,  12,  16. 
What   is   the   length    of    ;i    diagonal?      What    is    its    length    if    Ilio 
dimensions  are  V&7  v!>.    \'lx> 

4.  Find  the  lot:! I  surface  area  of  a  regular  triangular  prism 
with  basal  edges  of  4  ft.  and  latorn!   edges  of  x  ft. 


PRISMS 


285 


PROPOSITION  X. 

559.  THEOREM.  Any  parallelepiped  is  equal  t(> 
a  rectangular  parallelopiped  having  an  equal  base 
and  altitude. 


-L''  /          B'^       / 

t  -    --/—-,  / 


Given   a   parallelopiped   P  with  base  ABC  and  alti- 
tude h. 

To  Prove  P  — a  rectangular  parallelopiped  H  having 
a  base  EFG  =  ABC  and  altitude  h. 

Proof.  SUG.  1.  Extend  edge  BC  of  P  and  the  three 
edges  parallel  to  BC  and  at  some  convenient  point 
on  line  BC  take  B'C'  =BC.  Through  B'  and  C" 
pass  planes  perpendicular  to  line  B'C',  forming 
the  parallelopiped  X  with  base  A' B'C'.  Extend 
the  edge  A  'B'  and  the  three  edges  parallel  to 
A'B'  and  at  some  convenieut  point  take 
EF  =  A'B'.  Through  E  and  F  pass  planes  per- 
pendicular to  EF  forming  the  parallelopiped  H. 
2.  Show  that  N  is  a  right  parallelo- 


286  SOLID  GEOMETRY 

piped  and  H  is  a  rectangular  parallelepiped. 
Show  that  P,  N,  and  H  have  the  same  altitude 
and  equal  bases. 

3.  Compare  P  and   N.     §§  548  IV,  541. 
Compare  N  and  H. 

4.  Compare  P  and  H. 
Therefore — 

560.  COR.   I.     The  volume  of  any  parallelopiped  is 
equal  to  the  product  of  its  three  dimensions. 

Given  a  parallelopiped  P  with  dimensions  a,  b,  c. 

To  Prove  Vol.  P  =  abc. 

Proof.  P  equals  a  rectangular  parallelopiped  with 
dimensions  a,  b,  c.  Why?  The  volume  of  this  second 
parallelopiped  is  abc. 

Therefore— 

561.  COR.     II.     Two     parallelopipeds     u'ith     equal 
bases  are  to  each  other  as  their  altitudes. 

Proof.  Let  the  dimensions  be  a,  b,  c  and  a,  b,  c' , 
respectively.  Use  §  560  to  find  the  ratio. 

562.  COR.  III.     Two  parallelopipeds  with  equal  alti- 
tudes are  to  each  other  as  their  bases. 

Proof.  Let  the  dimensions  be  a,  b,  c,  and  a,  b ' ,  c' 
respectively.  Use  §  560  to  find  the  ratio. 

563.  COR.    IV.     Two    parallelopipeds    are    to    each 
other  as  the  products  of  their  three  dimensions. 

1.  Find  the  lateral  area  of  a  regular  pentagonal  prism  each 
edge  of  which  is  3  in. 

2.  If  a  secant  plane  intersects  two  planes  so  that  the  lines 
of  intersection  are   parallel   and   the   corresponding   dihedrals   are 
equal,  the  two  planes  are  parallel. 

3.  Prove   the   preceding  example   when    the   equal   dihedrals 
are  the  alternate  interior  angles;  prove  it  for  equal  alternate  ex- 
terior angles. 


PRISMS  287 

PROPOSITION  XI. 

564.  THEOREM.  The  volume  of  a  triangular 
prism  is  equal  to  the  product  of  its  base  and  its 
altitude. 


F 

Given  the  triangular  prism  EFG  —  N,  denoting  its 
volume  by  V,  its  base  by  B,  and  its  altitude  by  h. 

To  Prove  V  =  hB. 

Proof.  SUG.  1.  Extend  the  planes  of  the  bases. 
Through  the  edges  AE  and  CG  pass  planes  paral- 
lel to  the  faces  NG  and  NE  respectively.  The 
figure  DF  is  a  parallelepiped.  Why  ? 

2.  What  is  the  volume  of  DF  in  terms 
of  B  and  h /    Why? 

3.  What  is  the  volume  of  the  A-prism 
in  terms  of  DF  ?    §  550. 

4.  What  is  the  volume  of  the  A-prism 
in  terms  of  B.  and  h  ? 

Therefore— 

1.  Compare   the  volumes  of  two   rectangular  parallelepipeds 
the  respective  edges  of  which  are  2',  3',  7'  and  5',  3/,  8'. 

2.  A  parallelepiped  has  an  altitude  of  8  in.  and  its  base  is 
a,  rhombus  with  a   10  in.  side,  the  shorter  diagonal  being   12  in. 
Find  the  volume. 


288 


SOLID  GEOMETRY 
PROPOSITION   XII. 


565.     THEOREM.     The  volume  of  any  prism  is 
equal  to  the  product  of  its  base  and  its  attitude. 


Given  the  prism  GD  with  V,  B,  and  h  denoting  its 
volume,  base,  and  altitude  respectively. 
To  Prove  V  =  hB. 

Proof.  SUG.  1.  Through  any  one  lateral  edge  pass 
diagonal  planes.  Into  what  kind  of  figures  is  the 
prism  divided? 

2.     Denote  the   respective   volumes   and 
bases  of  these  figures  by  Vlf  Bl;  V2,  B     F 
etc.    Then 


8, 


V3  =  h  X  B3,  etc.     Why? 
3.    V=Vl+V2+V8.... 


4.     Determine  V  in  terms  of  h  and  B. 
Therefore— 

566.  COR.  I.     //  two  prisms  have  equal  bases,  their 
volumes  are  proportional  to  their  altitudes. 

567.  COR.  II.     //   two  prisms   have   equal   altitudes 
their  volumes  are  proportional  to  their  bases. 


EXERCISES  280 

1.  Find  the  volume  of  a  regular  hexagonal  prism,  the  lateral 
edge  being  10  in.  and  the  basal  edge  being  5  in. 

2.  Find    the    volume    of    a    rectangular    parallelepiped,    the 
lateral  edge  being  20'  and  the  basal  edges  being    5'. 

3.  Find  the  volume  of  a  parallelepiped,   the  base  being  8" 
square,  the  lateral  edge  13"  and  the  perpendicular  let  fall  from  a 
vertex   of   one   base   striking   the   other    base   5"    from   the    corre- 
sponding vertex. 

4.  Find   the   volume  of   a  hexagonal   prism,   having   for   its 
base   a    regular    polygon,    the    lateral   edge    being   25",    the   basal 
edge  10",  and  the  inclination  of  the  lateral  edge  to  the  base  be- 
ing 60°.     Find  the  volume  if  the  inclination  is  45°. 

5.  The  specific  gravity  of  iron  is  7.  4.       What  is  the  weight 
of   a   rectangular   tank    including   the    cover   \"   thick,     made    of 
iron    the   inside    measurements    being    20"  X  20"  X  4'?        Make  a 
diagram  showing  how  the  answer  can  be  obtained  with  the  least 
computation. 

6.  A   right   triangular   tank  has   a   lateral   edge   of   15'   and 
basal  edges  of  8',  7',  and  5',  inside  measurements.     Find  its  total 
area  and  its  capacity  in  gallons.     7J  gal.  =  1  cu.  ft.  approximately. 

7.  How  can  one  obtain  the  volume  of  an  irregular  piece  of 
rock  by  immersing  it? 

8.  What   is  the  ratio   of  two   rectangular  solids  the   dimen- 
sions of  which  are  3,  8,  12  and  4,  9,  20  respectively? 

9.  The   apothem    of    a   regular    hexagonal   prism   is    10    and 
its  lateral  edge  is  20.     Find  its  volume  and  total  area. 

10.  The  apothem  of  a  cube  is  4.  Find  volume  and  total  area. 

11.  The  edges  of  three  cubes  are  7",  12",  and  13"  respectively. 
Find  the  volume  of  each  and  the  total  area  of  each. 

12.  A    rectangular    box    12"  X  18"  X  22",    outside    measure- 
ment, is  made  of  1"  boards.    What  is  its  capacity  in  cubic  inches? 
How  many  cubical  boxes     2f"   on  an  edge  can  be  packed  in  it? 
What  is  its  capacity  in  gallons? 

13.  A  cistern  is  in  the  form  of  a  regular  hexagonal  prism. 
The  lateral  edge  is  7'  and  the  basal  edge  is  6',  inside  measure- 
ments.    What  is  its  capacity  in  gallons? 

14.  The  volume  of  a  rectangular  parallelepiped  is  6,720  cubic 
inches  and  its  edges  are  in  the  ratio  of  3,  5,  7.     Find  the  three 


290  SOLID  GEOMETRY 

568.  PYRAMID.     A   polyhedron   all   but   one   of   the 
faces  of  which  meet  in  the  same  point  is  a  pyramid.   The 
point  in  which  these  faces  meet  is  the 

vertex.  The  face  which  does  not  meet 
the  vertex  is  the  base  and  the  other  faces 
are  the  lateral  faces.  The  intersections 
of  the  lateral  faces  are  the  lateral  edges D< 
and  the  intersections  of  the  lateral  faces 
with  the  base  are  the  basal  edges. 

Point  out  the  various  parts  of  the  pyramid  above. 

569.  ALTITUDE  OF  A  PYRAMID.     The  length  of  the 
perpendicular  from  the  vertex  to  the  plane  of  the  base 
is  the  altitude  of  the  pyramid.    It  may  also  be  taken  as 
the  perpendicular  distance  between  the  plane  of  the  base 
and  a  plane  through  the  vertex  parallel  to  the  base. 
Why? 

570.  PYRAMIDS  CLASSIFIED.     A  pyramid  is  triangu- 
lar, quadrangular,  pentagonal,  etc.,  according  as  its  base 
is  a  triangle,  quadrilateral,  a  pentagon,  etc. 

In  a  triangular  pyramid  any  face  may  be  taken  for  the  base, 
the  vertex  of  the  opposite  polyhedral  angle  then  being  the  ver- 
tex of  the  pyramid. 

571.  REGULAR  PYRAMID.     A  pyramid  with  a  regular 
base,    such   that   the   vertex  lies   in   the   perpendicular 
erected  at  the  center  of  the  base,  is  a  regular  pyramid. 

572.  SLANT   HEIGHT.     The  perpendicular  from  the 
vertex  of  a  regular  pyramid  to  any  basal  edge  is  the 
slant  height  of  the  regular  pyramid. 

573.  TRUNCATED  PYRAMID.     That  portion  of  a  pyra- 
mid included  between  the  base  and  a  plane  cutting  all 
the  lateral  edges  is  a  truncated  pyramid. 

574.  FRUSTUM  OF  A  PYRAMID.      A  truncated  pyramid 


PYRAMIDS 


291 


iii  which  the  cutting  plane  is  parallel  to  the  base  is  a 
frustum  of  a  pyramid.  The  section  made  by  the  cutting 
plane  is  the  upper  base  of  the  frustum. 

575.  ALTITUDE  OF  A  FRUSTUM.      The  perpendicular 
distance  between  the  bases  is  the  altitude  of  the  frus- 
tum. 

COROLLARIES  TO  THE  DEFINITIONS. 

576.  COR.  I.     The  lateral  faces  of  a  pyramid  are  tri- 
angles. 

577.  COR.  II.     In  a  regular  pyramid  the  lateral  edges 
are  equal,  the  lateral  faces  are  congruent  triangles,  arid 
the  slant  height  is  the  same  irrespective  of  the  face  in 
uHiich  it  is  drawn. 

578.  COR.  III.      In  a  frustum  of  a  regular  pyramid 
the  lateral  edges  are  equal,  the  lateral  faces  are  con- 
gruent trapezoids,  and  the  slant  height  is  the  same  irre- 
spective of  the  face  in  ivhich  it  is  drawn. 


PROPOSITION   XIII. 

579.     THEOREM.     //  a  pyramid  is  cut  by  a  plane 
parallel  to  the  base, 

I.     The  lateral  edges  and  the  altitude  are  cut 
proportionally; 


292  SOLID  GEOMETRY 

II.     The  section  is  a  polygon  similar  to  the  base. 

Given  a  pyramid  P  —  AC  with  base  ABC...  cut  by 
a  plane  M  II  the  base  in  the  section  A'B'C' .  .  .  with  alti- 
tude PO. 

_    _          PA      PB      PC  PO 

I.  To  Prove = = ,  etc..  =  — 

PA'     PB'     'PC'  PO' 

Proof.     SUG.  1.     Through   the  vertex  P  pass   plane 
N  II  plane  M. 

2.     Compare  the  ratios 
PA      PB      PC      PO  9 

—        •      —        '     —        •      — •       <o    "io^J. 

PA''   PB''    PC'9   PO' 

II.  To  Prove  A'B'C'  ^  ABC  . . . 

Proof.     SUG.  1.     What   is   the    definition   of   similar 
polygons  ? 


2 = = ,  etc. 

A'B'      PB'     B'C' 

3.     Complete  the  demonstration. 
Therefore— 

580.  COR.       The  bases  of  a  frustum  of  a  pyramid 
are  similar  polygons. 

PKOPOSITION  XIV. 

581.  THEOREM.     //  a  section  of  a  pyramid  is 
parallel  to  the  base,  the  ratio  of  the  section  to  the 
base  equals  the  ratio  of  the  squares  of  the  dis- 
tances from  the  vertex  to  the  section  and  the  base. 

Given  a  pyramid,  as  in  §  579,  with  section  A'B'C' 
parallel  to  the  base  ABC. .  .  and  with  PO'  and  PO  the 
respective  distances  of  the  section  and  the  base  from 
the  vertex  P. 

A'B'C'...     PO'2 

T°PrOVe          1^7      ^ 


PYRAMIDS 


293 


Proof.     SUG. 


ABC  ...  ZB2  '"  PA2~  PO'2' 
Give  the  authority  for  each  of  these  statements. 
Therefore — 

1.  The  three  edges  of  a  rectangular  parallelepiped  meeting 
in  a  point  are  2',  5',  and  10'.     Find  its  lateral  area  and  volume. 

2.  A  right  triangular  prism  has  an  altitude  of  20"  and  basal 
edges    of   10',    19',    and    ]2'.      Find   its   lateral    area    and    volume. 
Two  methods. 

PROPOSITION  XV. 

582.  THEOREM.  In  pyramids  having  equal 
bases  and  equal  altitudes,  sections  made  by  planes 
parallel  to  the  respective  bases  and  at  equal  dis- 
tances from  the  respective  vertices  are  equal. 


Given  two  pyramids,  P  and  P',  with  equal  altitudes 
h,  equal  bases  B  and  B'  respectively;  M  and  M'  being 
sections  of  P  and  P'  parallel  to  the  respective  bases  and 
at  the  same  distance  d  from  the  respective  vertices. 

To  Prove  M=M'. 


. 
Proof.     SUG.  1. 

Therefore — 


M     c/2       ,  M'     c/2      ™ 
—  =  —  and  — = —     Why? 
B      7i*  B'       7,2 


294 


SOLID  GEOMETRY 


PROPOSITION  XVI. 

583.  THEOREM.     The  lateral  area  of  a  regular 
pyramid  is  equal  to  one-half  the  product  of  the 
perimeter  of  the  base  and  the  slant  height. 

Given  the  regular  pyramid  0,  its 
lateral  area  denoted  by  S,  its  slant  height 
ON  by  I,  and  the  perimeter  of  its  base 
by  p. 

To  Prove  S  =  %lp. 

Proof.     Proof  left  to  the  student. 

^B  c 

PROPOSITION   XVII. 

584.  THEOREM.     The  lateral  area  of  the  frus- 
tum of  a  regular  pyramid  is  equal  to  the  product 
of  the  slant  height  by  one-half  the  sum  of  the 
perimeters  of  the  bases. 


Given  EG'  the  frustum  of  a  regular  pyramid,  its 
lateral  area  denoted  by  S,  its  slant  height,  NN',  by  I, 
and  the  respective  perimeters  of  the  two  bases  by  p 
and  pr . 

To  Prove  8  =  \  x  l(p  +//). 


PYRAMIDS 


295 


Proof.     SUG.  1.     Find  the  area  of  each  face  and  add 
these  areas. 

2.     What  is  the  coefficient  in  the  addi- 
tion ?    §  535. 

Therefore — 

585.  INSCRIBED  PRISMS.  If  a  triangular  pyramid 
is  cut  by  any  number  of  planes  parallel  to  its  base  and 
through  the  lines  in  which  these  planes  cut  one  of  the 
lateral  faces  of  the  pyramid  other  planes  are  passed 
parallel  to  the  opposite  lateral  edge  of  the  pyramid, 
certain  triangular  prisms  are  formed.  These  are  in- 
scribed prisms,  as  A,  B,  Ct  etc.,  or  circumscribed  prisms, 
as  A'B'C',  etc.,  according  as  this  second  set  of  planes 
meet  the  successive  planes  of  the  first  set  within  the 
pyramid  or  without  the  pyramid. 


586.  POSTULATE.  A  pyramid  has  a  volume  greater 
than  the  combined  volumes  of  any  set  of  inscribed 
prisms  and  less  than  any  set  of  circumscribed  prisms. 

1.  The   diagonal  of  a  rectangular  parallelepiped  is  24  and 
its  edges  are  in  the  ratio  of  2,  3,  4.     Find  the  total  area  and  the 
volume. 

2.  A  cubical  cistern  holds  900  bbls.  of  31*  gals.     How  many 
square  feet  of  surface  has  it? 

3.  Find   the   interior   surface    of   a  rectangular   cistern,   the 
edges  being  in  the  ratio  of  3,  4,  5  and  its  capacity  900  bbls. 


296 


SOLID  GEOMETRY 


PROPOSITION  XVIII. 

587.  THEOKEM.  The  volumes  of  two  triangular 
pyramids  with  equal  altitudes  and  equal  bases  are 
equal. 


D' 


Given  two  triangular  pyramids  P  and  P',  with  equal 
altitudes  h  and  equal  bases. 

To  Prove  P  =  P'. 

Proof.  SUG.  1.  Divide  the  equal  altitudes  of  P  and 
P'  into  any  number  of  equal  parts  x  and  through 
the  points  of  division  pass  planes  parallel  to  the 
bases  and  form  in  P  a  set  of  inscribed  prisms, 
A,  B,  C. . .  and  in  P'  a  set  of  circumscribed 
prisms  A',  B',  C',... 

2.  Prisms  B'  =  A,  C'  =  B,  #'=0,  etc. 
Why?     §  564. 

3.  Denote  A  +  B  +  C  + . .  .   by  V  and 
A'+B'  +  C'  +...    by  V.     Then  V  —  V  =  A'. 
Why? 

4.  Suppose  that  P'  >  P,  i.  e.  P'  —P 
equals    some    definite    number    K.      By    §    586 
P'  <  V  and  P  >  V  so  that  P'-  P  <  V'-V.    §  49. 

5.  Hence   P' —  P  <  A' .     Why?      By 


PYRAMIDS 


297 


taking  the  length  x  small  enough  the  prism  A' 
can  be  made  as  small  as  is  desired,  even  less 
than  K,  since  its  altitude  x  is  decreased.  §  564. 
Hence  P'  cannot  be  greater  than  P. 

6.     Form    the    inscribed   prisms    in    P' 
and  the  circumscribed  prisms  in  P.     It  can  now 
be  proved  that  P'  cannot  be  less  than  P. 
Therefore — 

PROPOSITION    XIX. 

588.  THEOREM.  The  volume  of  a  triangular 
pyramid  is  one-third  the  product  of  its  base  and 
its  altitude.  A  n- 

7 


Given  the  triangular  pyramid  A  —  ECD,  its  volume 
denoted  by  V,  its  base  by  B,  and  its  altitude  by  h. 

To  Prove  V=   $hxB. 

Proof.  SUG.  1.  Through  A  pass  a  plane  parallel  to 
the  base,  extend  the  planes  of  the  faces  ACE 
and  ACD,  and  through  ED  pass  a  plane  parallel 
to  AC.  The  resulting  figure  is  a  prism.  Why? 

2.  A  —  EE'D'D    is    a    quadrangular 
The  plane  of  AE'D  divides  it  into  two 

pyramids,    A  —  EE'D    and    A  —  E'D'D. 

3.  A—ECD  =  D—E'D'A  =  A—E'DE 

4.  A  —  ECD  =  J  prism.     What  is  its 
base  and  altitude  ?     Authority  for  each  statement. 


prism, 
equal 


298 


SOLID  GEOMETRY 
6.     Complete  the  demonstration. 


Therefore— 

PROPOSITION  XX. 

589.     THEOREM.    The  volume  of  any  pyramid  is 
equal  to  the  product  of  its  base  and  its  altitude. 


Given  pyramid  0,  denoting  its  volume  by  I7,  its  base 
by  B,  and  its  altitude  by  h. 

To  Prove  V  =  J  h  x  B. 

Proof.  The  proof,  similar  to  that  of  §  565,  is  left  to 
the  pupil. 

590.  COR.  I.     //  two  pyramids  have  equal  bases  then- 
volumes  have  the  same  ratio  as  their  altitudes. 

591.  COB.  II.     //  two  pyramids  have  the  same  alti- 
tude, their  volumes  have  the  same  ratio  as  their  bases. 

COR.  III.  Any  two  pyramids  are  proportional  to  the 
products  of  their  bases  and  altitudes. 

PROPOSITION    XXI. 

592.  THEOREM.     The  volume  of  a  frustum  of  a 
triangular  pyramid  is  equal  to  the  sum  of  the  vol- 
umes of  three  triangular  pyramids  each  with  the 
altitude  of  the  frustum  and  with  bases  equal  rr 


PYRAMIDS 


299 


spectively  to  the  upper  base  of  the  frustum,  the 
lower  base  of  the  frustum,  and  a  mean  propor- 
tional to  the  bases  of  the  frustum. 


Given  frustum  of  a  triangular  pyramid  F,  with  upper 
base  Bl  and  lower  base  B2. 
To  Prove  F  =  $  Ji(Bl  +  B2  +  ^B~B~2). 
Proof.     SUG.  1.     By  a  plane  through  GE'F'  cut  off 

a  triangular  pyramid  P,  with  base  Bl  and  altitude 

h.    What  is  its  volume  ? 

2.  By  a  plane  through  GEF'   cut  off 
a  triangular  pyramid  P8  with  base  B%  and  alti- 
tude h.    What  is  its  volume  ? 

3.  The  remaining  portion  is  a  triangu- 
lar pyramid,  P3. 

»  A/H^__^fJ^_    Vff^         Wh.? 

/xt^ff.   f,"          /n"rv 


Why? 


&GE'G'      G'E 


Therefore — 


300 


SOLID  GEOMETRY 


PROPOSITION   XXII. 

593.  THEOREM.  The  volume  of  the  frustum  of 
any  pyramid  is  equal  to  the  sum  of  the  volumes 
of  three  pyramids  each  with  the  altitude  of  the 
frustum  and  with  bases  equal  respectively  to  the 
upper  base  of  the  frustum,  the  lower  base  of  the 
frustum,  and  a  mean  proportional  to  the  two 
bases  of  the  frustum. 


Given  the  frustum  F  with  upper  base  B19  lower  base 
B2,  and  altitude  h. 

To  Prove    F  =  %  h  (J?x+  58+  vB~B~2). 

Proof.  SUG.  1.  The  lateral  edges  of  F  will  if  ex- 
tended, meet  in  a  point  forming  a  pyramid  P. 
Why?  Construct  a  triangular  pyramid  P'  with 
the  same  altitude  a  as  P  and  a  base  B2'  equal  to 
the  base  B2  of  P.  Cut  from  Pf  a  frustum  F' 
with  altitude-  h.  ' 

2.     Compare  the  volumes  of  P  and  P'. 
Auth. 

3.  Compare  #/  and  £,.     §  582. 

4.  The    pyramids    with    bases   B^    and 
B/and  altitudes  a  —  ft  are  equal.     Why? 


PYRAMIDS  301 

5.     Hence  the  frustum  F  =  F'.     Why? 
6.     F'=$h(B1'+B2'+^B~'B./)  Why? 


Therefore — 

1.  A  monument  is  25'  high,  18"  square  at  one  end,  30"  square 
at  the  other,  and  of  uniform  shape.  What  is  its  volume  in  cubic 
feet?  If  its  specific  gravity  is  7-J  what  does  it  weigh  in  tons? 

PBOPOSITION  XXIII. 

594.  THEOREM.  Two  tetrahedrons  having  a 
trihedral  angle  of  one  equal  to  a  trihedral  angle 
of  the  other  have  the  same  ratio  as  the  products 
of  the  three  edges  including  the  equal  trihedral 
angles. 


C' 

Given  two  tetrahedrons  0  —  ABC  and  0' A'B'C' 

with  equal  trihedrals  0  and  0'. 

To  Prove     °~ABC   -    ^xQJ8xOC 
0' -A'B'C'      O'A'xO'B'xO'C' 

Proof.  SUG.  1.  Superimpose  trihedral  0'  upon  tri- 
hedral 0  and  from  points  A  and  A '  drop  perpen- 
diculars to  the  opposite  face  meeting  it  in  points 
M  and  M'  respectively. 

2      Then      °~ABC   =  AM  x  A  OBC 

0- A'B'C'     A'M'x&OB'C'' 
Why? 


302  SOLID  GEOMETRY 

t 

&OBC        OBxOC 

3.  Also  -  =  —  — .     Why? 

AOB'C"      OB'xOC' 

4.  Points  0,  M,  M'  are  in  a  straight 
line.     Why  ? 

5,AK=oA_.    Why? 

A'M'       OA' 

6.     Complete  the  demonstration. 
Therefore— 

595.  SIMILAR    POLYHEDRONS.      If    two    polyhedrons 
have  the  same  number  of  faces  similar  each  to  each  and 
similarly  placed  and  have  their  corresponding  polyhe- 
dral angles  equal,  the  polyhedrons  are  similar. 

The  equal  angles  and  the  lines  and  faces  in  the  two 
polyhedrons   which    are    similarly    situated   are    called 
homologous  angles,  lines,  and  faces,  respectively. 
PRELIMINARY  THEOREMS. 

596.  THEOREM  I.     Homologous  lines  in  similar  poly- 
hedrons are  proportional. 

597.  THEOREM  II.     Homologous    faces     in     similar 
polyhedrons  are  proportional  to   the  squares  of  homo- 
logous lines. 

598.  THEOREM  III.     The   surfaces   of   similar   poly- 
hedrons are  proportional  to  the  squares  of  homologous 
lines.    §374. 

1.  Find   the    volume   of   a   regular   triangular   pyramid  with 
basal  edge  of  4  ft.  and  altitude  of  5  V~3  ft. 

2.  The  point   of  meeting  of   the  three   medians   of  an   equi- 
lateral triangle  is  f       of  the  length  of  the  median  from  each  ver- 
tex.    Ex.  30,  p.  212. 

3.  The  edges  of  a  regular  tetrahedron  are  each  a  ft.     Find 
its  slant  height,  altitude,  base  area,  lateral  area,  total  area,  and 
vc  lume. 

4.  Find   the   volume   of   a   regular   hexagonal   prism    with    a 
radius  of  2  ft.  and  lateral  edge  of  2  yards. 


PYRAMIDS  303 

PROPOSITION  XXIV. 

599.     THEOREM.     Two  similar  tetrahedrons  are 
proportional  to  the  cubes  of  homologous  edges. 


O' 


Given  two  similar  tetrahedrons,  0  and  0'  ,  edges  OA, 
OB,  OC    ...     being  homologous  to  O'A',  O'B',  O'C/ 
.  .    respectively. 

To  Prove   °=M±. 

0'      OA'* 

Proof.     SUG.  1.     Compare    the    trihedral    angles    0 


and  Q:  -  =  ^.x-.X--.  Why 

0'     O'A'xO'B'xO'C'     O'A'    O'B'    O'C' 


• 

f)  R  Of1 

3.     Substitute  for   -     -and-    -  in  (1), 
O'B'         O'C 

1.  Volumes  of  similar  solids  are  to  each  other  as  the  square 
roots  of  the  cubes  of  their  surfaces. 

2.  If  the  amount  of  lumber  used  is  to  be  the  same  in  either 
case  which  provides   the   greater  capacity,   one  barn  or  two,   the 
barns  being  similar  in  shape? 

3.  If  the  dimensions  of  a  window  be  doubled,  by  what  fac- 
tor is  its  lighting  capacity  increased? 

4.  If  the  surface  of  a  box  be  trebled  in  building  a  second 
box  of  the  same  shape,  by  what  is  the  capacity  Increased? 

5.  What  increase  in  material  is  required  in  building  a  new 
grain  box  having  the   same  shape  as  the  old  one  but   of   double 
the  capacity? 


304 


SOLID  GEOMETRY 


PROPOSITION  XXV. 

600.  THEOREM.  Two  tetrahedrons  are  simi- 
lar if  three  faces  of  one  are  respectively  similar 
to  three  faces  of  the  other. 


ir, 


C' 


Given  two  tetrahedrons  0  and  0'  with  the  three  faces 
about  0  similar  to  the  three  faces  about  0 '  respectively. 
To  Prove  O^O'. 

Compare     trihedrals     0     and     0'. 


1. 


Proof.     SUG. 
Auth. 

2. 
A'B',  B'C' 


with 


3. 


Compare   lines  AB,   BC,   CA 
C' A'  respectively.    Auth. 
Compare     &     ABC    and   A'B'C'. 
Auth. 

4.  The    corresponding    trihedrals    and 
dihedrals  are  equal.    Why? 

5.  The    conditions    for    similarity    are 
fulfilled.    Why? 

Therefore— 

601.  COB.  Two  tetrahedrons  having  a  dihedral  of 
one  equal  to  a  dihedral  of  the  other  and  the  including 
faces  of  the  first  dihedral  similar  respectively  to  the  in- 
cluding faces  of  the  second  dihedral  and  similarly  placed 
are  similar. 

Proof.  SUG.  Prove  the  faces  opposite  the  equal 
dihedrals  similar. 


PYRAMIDS 
PROPOSITION  XXVI. 


305 


602.  THEOREM.  Two  similar  polyhedrons  can 
be  divided  into  tetrahedrons  similar  each  to  each 
and  similarly  placed. 


Given  P  and  P'  similar  polyhedrons. 

To  Prove  that  P  and  P'  can  be  divided  into  tetra- 
hedrons similar  each,  to  each  and  similarly  placed. 

Proof.  SUG.  1.  Through  any  two  corresponding 
vertices  as  0  and  0'  draw  in  each  of  the  includ- 
ing faces  all  possible  diagonals.  The  correspond- 
ing faces  of  P  and  P'  are  thus  divided  into  the 
same  number  of  triangles,  similar  each  to  each 
and  similarly  placed.  Why? 

2.  Let   OE  and   O'E'    be   two   corre- 
sponding edges  and  let  OF  and  OG  be  the  two 
diagonals   including   between   them   OE.      Simi- 
larly let  O'F'   and  O'G'  be  the  corresponding 
lines  of  P'.     Cut  from  P  and  P'  the  respective 
tetrahedrons  0  —  EFG  and  0'  —  E'F'G'.     By 
Sug.  (1)  kOEF'-' kO'E'F'  and 

A  OEG  <-'  A  O'E'G'.  Also  dihedral  OE  =  dihedral 
O'E'.  Why?  Hence  tetrahedron  0  —  EFG ^ 
tetrahedron  0'  —  E'F'G'.  §  601. 

3.  In  the  remaining  portions  of  P  and 
P'  the  new  face    A  FOG  and  A  EFG  are  similar 
respectively    to    F'O'G'     and     E'F'G'.     Why? 


306  SOLID  GEOMETRY 

Also  the  ratio  of  the  new  edges  OF,  OG,  FG  to 
O'F',    O'G',   F'G'    respectively,    being   equal   to 

OF 

- — ;  (Why?)  the  original  ratio  of  similitude. 
O'E' 

4.  Since  the  trihedral  angles  0  and  0 ' 
of  the  removed  tetrahedrons  are  equal,  so  are  the 
remaining  polyhedral  angles  0  and  0' . 

5.  Hence    the    remaining    polyhedrals 
are  similar.    Why? 

6.  Repeat  the  process  on  the  remaining 
portions  of  P  and  P' . 

Therefore— 

1.  Similar  polyhedrons  have  the  same  ratio  as  the  cubes  of 
homologous  edges. 

Sue.     According  to  the  demonstration  of  602,  the  poly- 
hedrons may   be   divided   into   tetrahedrons.     Let   them   be 
respectively  Pt  P/,  P2  P2'  etc.     Let  at  and  a/  be  homol 
ogous  edges  of  Pa  and  PI',  etc.     Then  by  599. 

— ^  =    *,  >   — *-f  =.  -&J-,  etc.     But  the  ratios  of  the  a  's  are 
"i        a±  s     PZ        az  3 

equal.      Why!     Hence  the  ratios  of  the  P's  are  equal  and 
Pt  +  P2  +  Pa  +  .  .  .  _  Pi_  _  a^. 
P/+  P2'+  P3'+  .  .  .      P/      fll'« 

603.  REGULAR  POLYHEDRON.  A  polyhedron  in  whicli 
all  the  faces  are  regular  congruent  polygons  and  in 
which  the  polyhedral  angles  are  equal  is  a  regular  poly- 
hedron. A  regular  polyhedron  must  be  convex. 

2.  The  perimeter  of  the  mid  section  of  a  frustum  of  a  pyra- 
mid equals  one-half  the  sum  of  the  perimeters  of  the  bases. 

3.  A  grain  bin  holds  100  bu.     Another  bin  has  each  dimen- 
sion   equal    to    twice    the    corresponding    dimension    of    the    first. 
What  is  its  capacity? 

4.  The  surfaces   of  two  similar  solids  are  to  each   other  as 
the  cube  roots  of  the  squares  of  their  volumes. 


REGULAR  POLYHEDRONS 


307 


PROPOSITION  XXVII. 

604.  THEOREM.     At    most    only    five    regular 
polyhedrons  can  be  formed. 

Proof.  SUG.  1.  Show  that  three,  four,  or  five  regu- 
lar triangles  can  be  so  used  as  to  form  a  convex 
polyhedral  angle. 

2.  Show  that   more  than   five   regular 
triangles  cannot  form  a  convex  polyhedral. 

3.  Make  a  similar  test  for  the  regular 
tetragon  or  square.    §  510. 

4.  How  many  regular  pentagons  may 
be  used?    Auth. 

5.  Show  that   no   regular   polygon   of 
more  than  five  sides  can  be  used  to  form  a  convex 
polyhedral. 

Therefore— 

605.  SCHOLIUM.     That  five  regular  polygons  can  all 
be  constructed  is  illustrated  as  follows :  Cut  the  patterns 
below  from  cardboard  and  cut  the  material  half  through 
along  the  dotted  lines.    Fold  each  so  as  to  form  a  poly- 
hedron, pasting  strips  of  paper  along  the  edges  to  keep 
them  in  shape.    It  is  possible  to  prove  that  these  figures 
can  actually  exist  by  purely  mathematical  reasoning. 


308  SOLID  GEOMETRY 

i 

1.  The  total  areas  of  regular    polyhedrons    have    the    same 
ratios  as  the   squares   of  their   altitudes   or  as   the   squares  of  any 
two  homologous  edges. 

2.  Find  the  volume  of  a  regular  quadrangular  pyramid  with 
basal  edge  of  8'  and  slant  height  of  5'. 

3.  The  volume  of  a  truncated  triangular  prism  is  equal  to 
the  sum  of  the  volumes  of  three  pyramids  having  the  base  of  the 
prism  as  a  common   base  and  the  vertices  of  the  inclined  section 
as  respective  vertices. 


To    PROVE    EFD  —  ABC  =  E  —  ABC  +  V  —  ABC  +  F  —  ABC. 

Pass  planes  through  each  vertex  of  the  truncating  section,  DEF, 
and  the  respective  opposite  edges  of  the  base.  Plane  EAC  cuts 
off  the  first  pyramid.  There  is  left  E-^ADCF.  Plane  EDC 
cuts  off  E  —  ACD  =  B  —  ABC  oi-D  —  ABC,  §570  the  second 
pyramid.  The  remaining  pyramid,  E  —  DCF  or  D — FEC  —  1)  — 
FCB  (equal  bases).  But  D  —  FCB  =  A—FCB,  the  third  pyramid 

4.  Find   the    volume   of   a  truncated   triangular   right    prism 
with  basal  edges  of  5',  5',  8'  and  lateral  edges  of  7',  8',  9'respeet- 
ively.     Use  two  methods  of  finding  the  base. 

5.  The    volume    of    any    truncated    triangular    prism    equals 
the  product  of  a  right  section  and  one-third  the  sum  of  the  lateral 
edges. 

6.  The   sides   of   a    right   section    of  a  truncated   triangular 
pyramid  are  8',  9',  15' and  the  lateral  edges  are  5',  9',  32' respect- 
ively.    Find  the  lateral  area  and  the  volume. 

7.  The  apothem  of  a  cube  is  7'.     Find  the  volume  by  using 
(he  rule  for  pyramids. 

8.  Find  the  volume  of  a  monolith  '24'  high,  J)'  square  at  one 
end  and  4'  square  at  the  other. 

Solve  without  pencil. 

9.  Find  the  slant   height,  the   apothem,   the   lateral   surl'.-icc, 
the  total  surface,  and  the   volume   of  a  regular  tetrahedron  with 
an  edge  of  5. 


CHAPTER  IX. 


THE  THREE  ROUND  BODIES. 

606.  CYLINDRICAL  SURFACE.    A  surface  formed  by  a 
moving  straight  line  which  always  remains  parallel  to 
its  original  position  and  continually 

touches  a  fixed  curved  line  is  a 
cylindrical  surface.  The  moving 
straight  line  is  the  generatrix  and 
the  fixed  curve  is  the  directrix.  The 
generatrix  in  any  one  of  its  posi- 
tions is  an  element  of  the  surface. 
If  the  directrix  is  a  closed  curve, 
the  cylindrical  surface  is  a  closed  cylindrical  surface. 

607.  A  CYLINDER.    A  solid  bounded  by  a  closed  cylin- 
drical surface  and  two  parallel  planes  cutting  the  ele- 
ments is  a  cylinder.    The  plane  sur- 
faces are  the  bases  of  the  cylinder 

and  the  cylindrical  surface  is  the 
lateral  surface  of  the  cylinder.  The 
distance  between  the  two  bases  is 
the  altitude  of  the  cvlinder. 


-  608.     RIGHT  SECTION.     A  section  of  a  cylinder  made 

by  a  plane  perpendicular  to  an  element  is  a  right  section. 

609.     CIRCULAR  CYLINDER.     A  cylinder  the  bases  of 

which  are  circles  is  a  circular  cylinder.    The  line  joining 

the  centers  of  the  bases  is  the  axis  of  the  cylinder. 


310  SOLID  GEOMETRY 

i 

610.  RIGHT  CYLINDER.    A  cylinder  the  bases  of  which 
are  perpendicular  to  the  elements  is  a  right  cylinder. 

611.  OBLIQUE  CYLINDER.     A  cylinder  in  which  the 
elements  are  not  perpendicular  to  the  bases  is  an  oblique 
cylinder. 

612.  CYLINDER  OF  REVOLUTION.    A  cylinder  generated 
by  the  revolution  of  a  rectangle  about  one  of  its  sides  is 
a  cylinder  of  revolution. 

613.  SIMILAR  CYLINDERS.    Cylinders  generated  by  the 
revolution  of  similar  rectangles  about  homologous  sides 
are  similar  cylinders. 

614.  TANGENTS  TO  A  CYLINDER. 
If  a  line  touches  the  lateral  surface 
of  a  cylinder  but  does  not  intersect 
it,  it  is  tangent  to  the  cylinder.    If 
a  plane  embraces  an  element  of  a 
cylinder  but  does  not  intersect  the 
surface,  it  is   tangent  to  the  cyl- 
inder. 

EF  is  tangent  to  the  cylinder  at  0  and  plane  N  is  tangent  in 
the  element  GH. 

PRELIMINARY  THEOREMS. 

615.  THEOREM  I.     The   elements   of   a   cylinder   are 
parallel  and  equal. 

616.  THEOREM  II.     A  right  section  of  a  cylinder  is 
perpendicular  to  every  element. 

617.  THEOREM  III.     A   cylinder   of  revolution   is   a 
right  circular  cylinder. 

1.          Sect  a  is  perpendicular  to  sort  ?>.     Tf  ft  is  revolved  about 
its  free  extremity  in  one  plane  what   figure  is  formed  by  a .' 


CYLINDERS  311 

PROPOSITION   I. 

618.  THEOREM.  Every  section  of  a  cylinder 
made  by  a  plane  embracing  an  element  is  a  par- 
allelogram. 


Given  the  plane  AD  embracing  the 
element  AB. 

To  Prove  that  plane  AD  intersects  the 
surface  of  the  cylinder  in  ZZ7  ABCD. 


Proof.  SUG.  1.  Let  D  be  the  point  in  which  the 
plane  cuts  the  perimeter  of  the  base  the  second 
time  and  through  D  draw  the  element  DC.  How 
does  DC  lie  with  respect  to  BAt  Why? 

2.  DC  and  BA  determine  a  plane  con- 
taining BA  and  D.    "Why  ? 

3.  How  many  planes  can  contain  BA 
and  D  ?    Where  then  does  DC  lie  with  respect  to 
the  given  plane  ABDt     With  respect  to  the  in- 
tersection of  the  plane  and  the  cylinder? 

4.  Complete  the  proof  by  showing  that 


Therefore  — 

619.  COR.     A  plane  containing  an  element  of  a  cylin- 
drical surface  without  being  tangent  inter  seels  the  sur- 
face in  a  second  element  also. 

620.  COR.  II.  Every  section  of  a  right  cylinder  by 
a  plane  which  embraces  an  element  is  a  rectangle. 

].  Cut  a  cylinder  of  revolution  by  a  plane  parallel  to  an 
element  in  such  a  manner  that  the  section  shall  be  a  rectangle 
congruent  to  the  rectangle  which  generates  the  cylinder. 


312  SOLID  GEOMETRY      . 

PROPOSITION  II. 

621.     THEOREM.     The   bases  of  a  cylinder  are 
congruent. 


Given  a  cylinder  AD  with  bases  ABC  and  DEF. 

To  Prove  bases  ABC  and  DEF  congruent. 

Proof.  SUG.  1.  Let  A,  B,  C  be  any  three  points  in 
the  perimeter  of  one  base  and  draw  the  element 
AD.  Pass  planes  through  the  element  AD  and 
the  points  B  and  (7,  intersecting  the  cylinder  in 
the  elements  BE  and  CF  respectively.  ADEB, 
ADFC,  BCFE  are  /I7.  Why? 

2.  Compare       &  ABC      and       DEF. 
Auth. 

3.  Superimpose    A  ABC  upon    A  DEF. 
As  A,  B,  and  C  are  any  three  points  of  the  perim- 
eter ABC,  where  does  the  perimeter  ABC  fall? 

4.     Compare  the  two  bases. 
Therefore — 

622.  COR.  I.     Any  two  parallel  sections  cutting  the 
elements  of  a  cylinder  are  congruent. 

623.  COR.  II.     All  sections  of  a  cylinder  parallel  to 
the  bases  are  congruent  to  the  bases. 

624.  COR.  III.     The  axis  of  a  circular  cylinder  passes 
through  the  centers  of  all  sections  which  are  parallel  to 
the 


CYLINDERS 


313 


SUG.  Draw  any  two  diameters 
in  one  base.  Pass  planes  through 
these  two  diameters  and  the  ele- 
ments at  their  extremities.  Where 
will  these  two  planes  intersect  the 
section?  The  other  base?  Where 
does  the  axis  lie  with  respect  to 
these  planes  ? 

1.         A  plane  tangent  to  a   cylinder  of  revolution  is  perpen- 
dicular to  the  plane  of  any  right  section. 

625.  CYLINDER  INSCRIBED  IN  A 
PRISM.     If  each  lateral  face  of  a 
prism  is  tangent  to  a  cylinder  and 
the  bases  of  the  prism  are  circum- 
scribed   about    the    corresponding 
bases  of  the  cylinder,  the  prism  is 
circumscribed   about   the    cylinder 
and  the  cylinder  is  inscribed  in  the 
prism. 

626.  CYLINDER     CIRCUMSCRIBED 
ABOUT  A  PRISM.     If  each  lateral 
edge  of  a  prism  is  an  element  of  a 
cylinder  and  the  bases  of  the  prism 
are  inscribed  in  the  corresponding 
bases  of  the  cylinder,  the  prism  is 
inscribed  in  the  cylinder  and  the 
cylinder  is  circumscribed  about  the 
prism. 

627.  POSTULATE  I.     The  volume  of  a  circumscribed 
prism,  the  areas  and  perimeters  of  its  bases  and  sections, 
and  its  lateral  area  are  greater  than  the  corresponding 
parts  of  an  inscribed  cylinder. 


314  SOLID  GEOMETKY 

628.  POSTULATE  II.     The    volume    of    an    inscribed 
prism,  the  areas  and  perimeters  of  its  bases  and  sections, 
and  its  lateral  area  are  less  than  the  corresponding  parts 
of  a  circumscribed  cylinder. 

629.  POSTULATE  III.     //  the  number  of  sides  of  a 
prism  inscribed  in  a  cylinder  or  circumscribed  about  a 
cylinder  be  indefinitely  increased  in  such  a  manner  that 
the  sides  of  the  prism  be  all  indefinitely  decreased,  the 
volume,    lateral    surface,    bases,    sections,     perimeters 
of  bases  and  sections  of  the  cylinder  are  the  limits  of 
ihe  corresponding  parts  of  the  prism. 

PROPOSITION  III. 

630.  THEOREM.     The  area  of  the  lateral  sur- 
face of  a  cylinder  is  equal  to  the  perimeter  of  a 
right  section  multiplied  by  an  element  of  the  sur- 
face. 


Given  cylinder  AG,  its  lateral  area  denoted  by  S,  the 
perimeter  of  the  right  section  by  p,  and  the  element  AE 
by  e. 

To  Prove  S  =  p  x  e, 

Proof.  SUG.  1.  Inscribe  in  the  cylinder  a  prism, 
d3noting  its  lateral  area  by  S',  the  perimeter  of 
its  right  section  p  by  p'.  Its  edge  is  e.  Why? 

2.  Then  S'=p'  X  e.     Why? 

3.  Indefinitely  increase  the  number  of 


CYLINDERS  315 

faces  in  the  manner  indicated  in  §  629.    What  are 
the  limits  of  S',  p'  ,  and  p'  X  el 

4.     •'•S  =  p*e.    Why? 
Therefore  — 

631.  COB.  I.     The  lateral  area  of  a  cylinder  of  revo- 
lution  is   equal   to    the   product   of   the   circumference 
of     the     base     and     an     element,     or     the     altitude. 
i.  e.    S  =  2"rh. 

632.  COR.  II.     The  lateral  areas  of  similar  cylinders 
are  to  each  other  as  the  squares  of  their  altitudes  and  as 
the  squares  of  the  radii  (or  diameters)  of  their  bases. 

SUG.     Denote  the  lateral  areas  by  8±  and  S2, 
the  altitudes  by  h^  and  h2,  the  radii  by  ri  and  r2, 
the  diameters  by  dl  and  <22  respectively. 
Then  ^i  =27r^  fti  =»*i  fti  =  r,  x  fei  =  r,  2  =fe,  «  ==d1  ^ 
#2      2*r27i2      r27i2      r2      7i2      r22     7i22     d22' 
Give  the  reasons  for  each  statement. 

633.  COR.  III.     The  total  area,  A,  of  a  cylinder  of 
revolution,  which  is  the  sum  of  the  lateral  area  and  the 
areas  of  the  two  bases,  is  given  by  the  formula  A  =  2*rh 


634.  COR.  IV.  The  total  areas  of  two  similar  cylin- 
ders are  to  each  other  as  the  squares  of  their  altitudes, 
and  as  the  squares  of  their  radii  (or  diameters). 

Proof  left  to  the  student. 

1.  How  many  feet  of  lumber  will  it  take  to  build  the  walls 
of  a  circular  silo    15'    in  diameter  and  25'  high,   allowing    ^   for 
matching? 

2.  A  cistern  is   10'  deep   and   8'  in   diameter.      How  many 
square  yards  of  cement  is  needed  to  line  it? 

3.  The  total  area  of  a  right  circular  cylinder  is  SOvr  sq.  ft. 
and  the  radius  of  the  base  is  5  ft.  Find  the  altitude. 

4.  Which   generates  the   greater   lateral  area,  the  revolution 
of  a  rectangle  about  its  longer  side  or  about  its  shorter  side? 


316  SOLID  GEOMETRY 

t 

PROPOSITION  IV. 

635.     THEOREM.     The  volume  of  a  cylinder  is 
equal  to  the  product  of  its  base  and  its  altitude. 


Given  a  cylinder  Fft,  its  volume  denoted  by  V,  its 
base  by  B,  and  its  altiude  by  h. 
To  Prove  V  =  B*h. 
Proof.     SUG.  1.     Inscribe   in   the   cylinder  a   prism, 

its  volume  denoted  by  V  and  its  base  by  B' .    Its 

altitude  will  be  h.    Why  ? 

2.  Then  V'=B'  x  ft.     Why? 

3.  Indefinitely  increase  the  number  of 
sides  of  the  prism  in  the  manner  indicated  in  §  629. 
What  is  the  limit  of  V  ?    Of  IT  ?    Of  the  product 
fl'Xft?     Why? 

4.  Complete  the  demonstration. 
Therefore — 

636  COR.  I.  The  volume  of  a  cylinder  of  revolution 
is  expressed  in  terms  of  its  altitude  h  and  radius  r  by 
the  formula  V  =  7rr2h. 

637.  COR.  II.  The  volumes  of  two  similar  cylinders 
of  revolution  are  to  each  other  as  the  cubes  of  their  alti- 
tudes and  as  the  cubes  of  their  radii  (or  their  diam- 
eters). 

Proof  loft  to  the  pupil.    See  §  632. 


EXERCISES  317 

1.  What  must  be  the  shape  of  a  piece  of  paper  which  ex- 
actly  covers  the   lateral   surface    of   an   oblique   cylinder?      Of    a 
right   cylinder? 

2.  A    cistern   holding   75   bbls.   is    8    ft.   deep.     What   is   its 
diameter?     How   many   square   yards   of   cement  are   required   to 
line  it? 

3.  A  cylindrical  tank  on  a  water  works  tower  has  a  lateral 
area  of  1,232  sq.  ft.     The  radius  of  its  base  is     \     the  altitude. 
Find  the  altitude  and  the  radius.     Find  the  total  area.     What 
would  be  the  lateral  area  if  the  altitude  were  doubled.     Find  its 
total  area  under  the  same  condition. 

4.  What  is  the  locus  of  a  point  in  space  at  a  given  distance 
from  an  unlimited  straight  line? 

5.  A  saw  log  is  16'  long  and  18"  in  diameter  at  the  small 
end.     How  many   board   feet  in  the  largest   squared   timber   that 
tan  be  sawed  from  it? 

6.  A    cylindrical    water    reservoir   is    110'    in    diameter    and 
20'  deep.     How  many  bbls.  does  it  hold?     How  many  square  yards 
of  cement  are  necessary  to  line  it? 

7.  A  rock  submerged  in  a  tank  29'  in  diameter  and  8'  deep 
raised  the  water  2'.     What  was  the  volume  of  the  rock? 

8.  A  farmer  feeds  his  60  cows  2  bu.  of  ensilage  each  a  day. 
The    silo    is    cylindrical,    30'    in    diameter    and    40'    deep.      How 
long  will  the  ensilage  last? 

0.        There  are  wine  tanks   15'  deep  and  10'  in  diameter.  How 
many  quart  bottles  can  be  filled  from  one  of  them? 

10.  What  is  the  ratio  of  the  lateral  area  of  a  right  circular 
cylinder  to  the  sum  of  the  bases? 

11.  What  is  the  locus  of  a  sect  at  a  given  distance  from  a 
straight  line  to  which  it  is  parallel? 

12.  What  is  the  locus  of  point  X  which  is  at  a  given  distance 
from  a  straight  line  and  equally  distant  from  two  fixed  points? 

13.  What  is  the  locus  of  a  point  at  a  given   distance  from 
the  lateral  surface  of  a  cylinder?     Is  it  possible  for  one  element 
of  the  locus  to  lie  within  the  cylinder?     Under  what  condition? 

14.  From  a  given  point  without  a  cylinder  draw  a  plane  tan 
gent  to  the  cylinder. 


318  SOLID  GEOMETRY 

< 

REVIEW. 

638.  State  the  formula  for 

1.  The  lateral  area  of  a  cylinder  of  revolution. 

2.  The  total  area  of  a  cylinder  of  revolution. 

3.  The  volume  of  a  cylinder  of  revolution. 

•A.     The  ratio  of  the  lateral  area  of  two  similar  cylin- 
ders. 

5.  The  ratios  of  the  lateral  areas  of  any  two  cylinders 
of  revolution. 

6.  The  ratios  of  the  volumes  of  two  similar  cylinders. 

1.  A   protecting   wall   for   an  embankment   500'   long  is    12' 
wide  at  the  bottom,  3'  wide  at  the  top,  and  18'  high.    How  many 
cubic  yards  of  stone  in  the  wall? 

2.  How  many  brick,   2"  X  4"  X  8",  are  required  to  build  a 
chimney  two  bricks  thick  with  a  flue  8"  X  12"   and  25'  high,  if 
the  mortar  averages    i"  thick?        Make  a  drawing    of  two  tiers 
of  brick  when  properly  laid. 

3.  If  from  any  point  in  an  equilateral  triangle,  perpendicu- 
lars be  drawn  to  the  sides,  the  sum  of  these  three  perpendiculars 
equals  the  altitude  of  the  triangle. 

4.  If  from  any  point  within  a  regular  tetrahedron  perpen- 
diculars be  drawn  to  the  four  faces,  the  sum  of  these  perpendicu- 
lars equals  the  altitude  of  the  tetrahedron. 

CONES. 

639.  CONICAL  SURFACE.     A  curved    surface    formed 
by  a  moving  straight  line  which  passes  through  a  fixed 
point  and  continually  touches  a   fixed 

curve  is  a  conical  surface.  The  mov- 
ing straight  line  is  the  generatrix 
and  the  fixed  curve  is  the  directrix. 
The  generatrix  in  any  one  of  its  posi- 
tions is  an  element  of  the  surface.  Tho 
fixed  point  is  the  vertex  of  the  coni- 
cal surface.  Ft'  the  directrix  is  a  closed 


CONES  319 

curve,  the  conical  surface  is  a  closed  surface.  The  por- 
tions of  the  conical  surface  on  the  two  sides  of  the  ver- 
tex are  the  nappes  of  the  conical  surface  and  are 
designated  as  upper  and  lower  nappes. 

Usually  only  one  nappe  of  a  conical  surface  is  considered. 

640.  A  CONE.     A  solid  bounded  by  one  nappe  of  a 
closed  conical  surface  and  a  plane  cutting  all  the  ele- 
ments, is  a  cone.       The  plane 

section  is  the  base  of  the  cone, 

The  bounding  portion  of  the 

conical  surface  is  the  lateral 

surface  of  the  cone  and  the 

vertex  of  the  conical  surface  is  the  vertex  of  the  cone. 

The  portions  of     the   elements   of  the   conical  surface 

between  the  vertex  and     the  base  are    the  elements  of 

the  cone.        The  distance  from  the  vertex  to  the  plane 

of  the  base  is  the  altitude  of  the  cone. 

641.  CIRCULAR  CONE.    A  cone  with  a  circular  base  is 
a  circular  cone,  and  the  straight  line  joining  the  vertex 
to  the  center  of  the  base  is  the  axis  of  the  circular  cone. 

642.  RIGHT  CONE.    A  cone  such  that  the  axis  is  per- 
pendicular to  the  base  is  a  right  cone. 

643.  OBLIQUE  CONE.     A  cone  such  that  the  axis  is 
not  perpendicular  to  the  base  is  an  oblique  cone. 

644.  CONE  OF  REVOLUTION.    A  cone  generated  by  the 
revolution  of  a  right  triangle  about  one  leg  as  an  axis  is 
a  cone  of  revolution.    Any  element  of  a  cone  of  revolu- 
tion is  its  slant  height. 

645.  SIMILAR  CONES.     Cones  of  revolution  that  can 
be  generated  by  the  revolution  of  similar  right  triangles 
about  homologous  sides  are  similar  cones. 


320  SOLID  GEOMETRY 

646.  TANGENTS  TO  A  CONE.  A  line 
which  touches  the  conical  surface  in 
a  point  but  does  not  intersect  it  is  a 
tangent  line  to  the  cone.  A  plane 
which  embraces  an  element  but  does 
not  intersect  the  conical  surface  is  a 
tangent  plane  to  the  cone. 


Lane  a  is  tangent  to  the  cone  at  the  point  of  tangency  J>  and 
plane  N  is  tangent  to  the  cone  along  the  element  GH. 

647.  A  TRUNCATED  CONE.    That  portion  of  a  cone  in- 
cluded between  the  base  and  a  plane  cutting  all  the  ele- 
ments is  a  truncated  cone. 

648.  A  FRUSTUM  OF  A  CONE.    A  truncated  cone  such 
that  the  cutting  plane  is  parallel  to  the  base  is  a  frus- 
tum of  a  cone.    The  base  of  the  cone 

is  the  lower  base  of  the  frustum,  the 

section  is  the  upper  base,  the  distance 

between  the  two  parallel  planes  is  the 

altitude.    If  the  frustum  be  a  portion 

of  a  cone  of  revolution,  the  portion  of 

the  slant  height  of  the  cone  included  between  the  base 4 

of  the  frustum  is  the  slant  height  of  the  frustutr 

PRELIMINARY  THEOREMS. 

649.  THEOREM  I.    Every  cone  of  revolution  is  a  rigid 
circular  cone. 

650.  THEOREM  It.     Tlic  a.cis  of  a  cone  of  revolution 
is  its  altitude. 

651.  THEOREM  III.     All  elements  of  a  cone  of  r<  so- 
lution arc  equal  (i.  e.  the  slant  height  ris  constant  to  the 
hypotenuse  of  the  generating  triangle). 


CONES 


321 


1.  Two  circular  cylinders  have  the  same  altitude,  but  the 
volume  of  one  is  four  times  that  of  the  other.  Find  the  ratio 
of  their  radii. 

PROPOSITION  V. 

652.     THEOREM.     Every  section  of  a  cone  em- 
bracing the  vertex  is  a  triangle. 


Given  a  plane  embracing  the  vertex  A  and  intersect- 
ing the  perimeter  of  the  base  in  the  points  B  and  C. 

To  Prove  that  the  A  ABC  is  the  section  made  by  the 
plane. 

Proof.  SUG.  1.  The  plane  must  cut  the  perimeter 
in  two  points  as  B  and  C. 

2.  AB  and  AC  are  each  in  both  the 
plane  and  the  conical  surface. 

3.  AB  and  AC  are  intersections  of  the 
plane  and  the  conical  surface. 

4.  What   kinds   of  lines   are   AB   and 
AC? 

Therefore— 

653.  COR.  I.  A  plane  containing  an  element 
of  a  conical  surface  without  being  tangent  intersects  the 
surface  in  a  second  element  also. 

2.  In  each  of  two  right  circular  cylinders  the  altitude  is 
equal  to  the  diameter  and  the  volume  of  one  is  ^  that  of  the 
other.  Find  the  ratio  of  the  altitudes. 


322  SOLID  GEOMETKY 

i 

PROPOSITION  VI. 

654.  THEOREM.  Every  section  of  a  circular 
cone  made  by  a  plane  parallel  to  the  base  is  a 
circle. 


Given  F  —  OGH  a  circular  cone  with  base  0  —  GH 
and  a  section  .O'—G'H'  parallel  to  the  base,  0  '  ,  G',H' 
being  the  points  in  which  the  plane  intersects  the  ele- 
ments FG}  FH,  and  the  axis  FO  respectively. 

To  Prove  O'—G'H'  a  O. 

Proof.  SUG.  1.  0  is  the  center  of  the  base  and  G 
and  //  are  taken  as  any  two  points  on  the  perim- 
eter of  the  base.  The  figures  FG'GOO'  and 
FH'HOO'  are  plane  figures.  Why? 

2.     Compare   the    &FG'0'    and   FGO; 


O'C*'  WTJ' 

FH'O'  and  FHO;  ratios   -    -,  -    -  and  - 

FO     OG  Oil' 

Auth. 

3.  Compare  O'G'  and  O'H'. 

4.  .'.  Q'  _  OH  is  a  circle.    Why? 
Therefore  — 

655.     COR.    The  axis  of  a  circular  cone  passes  through 
ike  centers  of  all  sections  parallel  to  the  base. 

What  is  the  character  of  the  polygon  which  revolved  about 
one  of  its  sides  generates  a  frustum  of  a  cone  of  revolution? 


COXES 


323 


656.  CONE  INSCRIBED  IN  A   PYRA- 
MID.   If  the  vertex  of  a  pyramid  coin- 
cides with  the  vertex  of  a  cone  and 
the  base  of  the   pyramid   is   circum- 
scribed about  the  base  of  the  cone,  the 
pyramid   is   circumscribed   about   the 
cone  and  the  cone  is  inscribed  in  the 
pyramid 

657.  CONE   CIRCUMSCRIBED  ABOUT 
A  PYRAMID.     If  the  vertex  of  a  pyra- 
mid coincides   with   the   vertex   of   a 
cone  and  the  base  of  the  pyramid-  is 
inscribed  in  the  base  of  the  cone,  the 
pyramid  is  inscribed  in  the  cone  and 
the  cone  is  circumscribed    about    the 
pyramid. 

PRELIMINARY   THEOREMS  AND  POSTULATE. 

658.  THEOREM  I.     The  faces  of  a  pyramid    circum- 
scribed about  a  cone  are  tangent  to  the  cone. 

659.  THEOREM  II.    The  edges  of  a  pyramid  inscribed 
in  a  cone  are  elements  of  the  cone. 

660.  THEOREM  III.     The  slant  height  of    a    regular 
pyramid  circumscribed  about  a  right  circular  cone  equals 
the  slant  height  of  the  cone. 

661.  POSTULATE  I.     The  volume  of   a    circumscribed 
pyramid,  the  area  and  perimeter  of  its  base,  and  its 
lateral  area  are  greater  than  the  corresponding  parts 
of  an  inscribed  cone. 

662.  POSTULATE  II.     The    volume    of    an    inscribed 
pyramid,  the  area  and  perimeter  of  its  base,  and  its 
lateral  area  are  less  than  the  corresponding  parts  of  a 
circumscribed  cone. 


324  SOLID  GEOMETRY 

t 

663.  POSTULATE  III.     //  the  number  of  sides  of  a 
pyramid  inscribed  in  or  circumscribed  about  a  cone  be 
indefinitely  increased  in  such  a  manner  that  the  faces 
of  the  pyramid  be  all  indefinitely  decreased,  the  volume, 
lateral  surface,  base,  perimeters  of  the  base  and  sections 
of  the    cone    are  the    limits  of  the    corresponding  parts 
of  the  pyramid. 

State  one  kind  of  inscribed  or  circumscribed  prisms  by 
•which  the  two  conditions  underlying  the  limiting  process  in  Postu- 
late III  may  be  set  up  in  one  operation. 

PROPOSITION  VII. 

664.  THEOREM.     The  lateral  area  of  a  cone  of 
revolution  is  equal  to  one-half  the  product  of  the 
perimeter  of  its  base  and  its  slant  height. 


Given  C  —  BD  a  cone  of  revolution,  its  lateral  area 
denoted  by  S,  the  perimeter  of  its  base  by  p,  and  its 
slant  height  by  I. 
To  Prove  S=$pxl. 

Proof.  SUG.  1.  Circumscribe  about  the  cone  a  regu- 
lar pyramid,  with  lateral  area  denoted  by  S' , 
perimeter  of  the  base  by  p'.  The  slant  height  of 
the  pyramid  is  I.  Why  ? 

2.     Determine    the    lateral    area   S'   in 
terms  of  p'  and  I. 


CONES 


325 


3.     Complete    the    demonstration.      See 
the  method  of  §  630. 

Therefore— 

665.  COR.  I.    //  r  be  the  radius  of  the  base  of  a  cone 
of  revolution,  then  the  lateral  area  is  given  by  the  for- 
mula 8=\l  X  "2flfr=^rl. 

666.  COR.  II.     The  lateral  areas  of  two  similar  cones 
are  to  each  other  as  the  squares  of  their  altitudes  and  as 
the  squares  of  the  radii  (or  diameters)  of  their  bases. 

SUG.     See  method  of  §  632. 


667.  COR.  III.  The  lateral 
area  of  the  frustum  of  a  cone  of 
revolution  is  equal  to  one-half 
the  product  of  its  slant  height 
and  the  sum  of  the  perimeters  of 
its  bases. 


SUG.  If  a  regular  pyramid  be  circumscribed 
about  the  cone  from  which  the  frustum  is  cut,  the 
cutting  plane  will  cut  from  the  pyramid  a  frus- 
tum which  will  be  circumscribed  about  the  frus- 
tum of  the  cone.  Denote  the  radii  of  the  two 
bases  by  rl  and  r0.  the  lateral  area  by  S  and  the 
slant  height  by  ?.  Prove  that  8  is  equal  to 


668.  COR.  IV.  //  r  represent  the  rad- 
ius of  the  section  of  a  frustum  of  a  cone  of 
revolution  midway  between  the  bases,  prove 


SUG.     Show  r= 


326  SOLID  GEOMETRY 

« 

PROPOSITION    VIII. 

669.  THEOREM.  The  volume  of  a  cone  is  equal 
to  one-third  the  product  of  its  base  and  its  alti- 
tude. 


Given    the    cone  A  —  FD    with  altitude  h,  base  B, 
and  volume  V. 
To  Prove  V=%B  x  k. 

Sue.  1.  Inscribe  in  the  co^ie  a  pyramid,  de- 
noting its  base  by  Be ,  and  its  volume  by  V .  Its 
altitude  will  be  h.  Why  ? 

2.     Complete  the  proof  by  an  adaptation 
of  the  method  of  Prop.  VII. 
Therefore— 

670.  COR.  I.     The  volume  of  a  circular  cone  is  given 
by  the  formula  V=^rzh. 

671.  COR.  II.     The  volumes  of  similar  cones  are  to 
each  other  as  the  cubes  of  the  radii  of  their  bases,  as 
the  cubes  of  their  altitudes,  and  as  the  cubes  of  their 
slant  heights. 


EXERCISES  327 

SUG.    Adapt  the  method  of  §  666. 

672.  COR.  III.     The  volume  of  a  frustum  of  a  cone 
of  revolution  is  equal  to  one-third  the  product  of  the 
altitude  and  the  sum  of  the  upper  base,  the  lower  base, 
and  a  mean  proportional  between  the  two  bases,  i.  e. 
V=±h  (Bi+B2  +  ^'E^Ez)  in  which  B^  andB2  denote  the 
two  bases. 

SUG.  Inscribe  in  the  original  cone  a  pyramid. 
The  frustum  of  the  pyramid  made  by  the  cutting 
plane  of  the  cone  will  be  inscribed  in  the  frustum 
of  the  cone.  Proceed  as  in  Cor,  III,  vProp.  VII. 

673.  COB.  IV.    The  volume  of  a  frustum  of  a  cone  of 
revolution  is  given  by   the  formula  V=\*h  (rl*+rz*  + 
vTj'xT7;). 

REVIEW. 

674.  State  the  formulas  for 

1.  The  area  of  a  circle. 

2.  The  lateral  area  of  a  cone  of  revolution. 

3.  The   lateral  area   of   the   frustum   of   a    cone   of 
revolution. 

4.  The  volume  of  a  circular  cone. 

5.  The  ratio  of  the  volumes  of  two  similar  cones. 

6.  The  ratio  of  the  lateral  areas  of  two  similar  cones. 

1.  A  granite  monument  is  30'  high.     The  base  is  4'  in  diam- 
eter.    It  diminishes   gradually   in   size   to    a   cross   section   circle 
15"  in  diameter,  6'  from  the  top.     The  remainder  of  the  monu- 
ment is  conical.     Find  its  volume  in  cubic  yards  and  its  weight  in 
tons,  the  specific  gravity  being  2.65. 

2.  How  many  sq.  yds.  of  canvas  in  a  conical  tent  12$'  high 
»vith  a  base  diameter  of  12'? 

3.  Find  the  volume  of  the  solid  generated  by  the  revolution 
upon  one  of  its  sides  as  an  axis  of  an  equilateral  triangle  with 
an  edge  of  6'. 


328  SOLID  GEOMETRY 

4.  Find  the  ratios  of  the  volumes  generated  by  the  revolu- 
tion of  a  right  triangle  about   side  a,  about   side   6,   and   about 
hypotenuse  o  respectively. 

5.  The  altitude  of  the   frustum   of  a  cone  of   revolution  is 
£    the  altitude  of  the  cone.       What  is  the  ratio  of  their  volumes? 

6.  A  section  of   a  tetrahedron   cutting   four 
edges   at   their  mid  points  is  a  parallelogram. 

7.  A  grain  bin  is  6'   X  12'  X  20'.  How  many 
bushels  does  it  hold? 


8.  The  radii  of  the  two  bases  of  a  frustum  of  a  cone  are 
4'  and  9'  respectively.     Find  the  volume. 

Indicate  and  work   mentally. 

9.  What  is  the  locus  of  a  rod  12'  long,  one  end  being  sta- 
tionary on  the  ceiling  of  a  room  8'  high  and  the  other  end  resting 
on  the  floor? 

10.  Find  the  area  of  the  plane  figure  enclosed  on  the  floor 
in  the  preceding  exercise  and  the  volume  enclosed  by  the   floor 
and  the  locus. 

11.  A  tent  has  a  diameter  of  16'  and  a  vertical  side  wall  of 
5'.     The  covering  is  conical  in  shape  and  the  center  pole  is  12' 
high.     How  many  cubic  feet  of  air  does  the  tent  contain? 

12.  Find  the  lateral  area,  total  area,  altitude,  and  volume  of  the 
cone  that  can  be  circumscribed  about  a  regular  tetrahedron  with 
an  8"  edge. 

13.  A  cylinder  and  a  cone  of  revolution  have  the  same  altitude 
and  concentric  bases.     The  diameter  of  the  base  of  the  cone  is 
three  times  that  of  the  cylinder.    How  far  from  the  vertex  of  the 
cone  do  the  two  lateral  surfaces  intersect?     Suppose  the  diameter 
of  the  cylinder  to  be  f   that  of  the  cone,  where  do  the    surfaces 
intersect  ? 

14.  What  is  the  locus  of  lines  making  a  given  angle  with  a 
given  line  at  a  given  point  in  the  line  ? 

15.  The   altitude   of   a   cone   is   trisected   by   planes   parallel 
to  the  base.     Compare  the  parts  into  which  the  cone  is  divided. 
Make  a  similar  comparison  when  the  altitude  is  divided  into  four 
equal  parts. 

16.  What  kind  of  a  triangle  is  the  section  of  a  right  cone 
through  the  vertex? 


EXERCISES  320 

17.  What    is    the    volume    of    a    piece    of    timber    15'    long, 
the  bases  being  squares  of  12"  and  14"  respectively? 

18.  If  four  similar  cylinders  have  their  altitudes  proportional 
to  3,  4,  5,  6,  prove  that  the  volume  of  the  largest  one  equals  the 
sum  of  the  volumes  of  the  three  others. 

19.  A  log  20'  long  has  a  diameter  at  the  smaller  end  of  16". 
What  proportion  of  the  log  is  cut  into  slabs  if  the  largest  possi- 
ble squared  stick  of  timber  is  sawed  from  it?     What  proportion 
will  be  slabs  if  the  largest  rectangular  stick  be  sawed,  the  edges 
having  the  ratio  of  3  to  4?     Of  2  to  3? 

20.  A  cylindrical  vessel  with  a  diameter  equal  to  its  altitude 
holds  1414 f  cu.  ft.  of  water.     What  are  its  dimensions? 

21.  What  are  the  dimensions  of  a  cylindrical  vessel  holding 
1414f  cu.  ft.  if  its  altitude  is  twice  its  diameter?      What  are  the 
dimensions  if  the  ratio  of  the  diameter  and  the  altitude  is  f  ? 

22.  What  are  the  dimensions  of  a  quart  cup  if  the  ratio  of 
its  diameter  and  its  altitude  is  f? 

23.  What  is  the  diameter  of  a  cylindrical  peck  measure  8"" 
deep? 

24.  Allowing  two  inches  for  the  seam,  how  large  a  sheet  of 
iron  will  be  required  for  a  joint  of  8"  stove  pipe?     Of  6"  stove 
pipe? 

25.  Measure  the  diameter  of  a  quart   cup   and   compute  its 
depth.     Verify  by  measurement. 

26.  A  rectangle  is  5x8.  Which  side  as  an  axis  will  produce  the 
greater   cylinder   of   revolution?     What   is   the   ratio   of   the   two 
cylinders? 

27.  Find  the  lateral  area  of  a   regular  pentagonal  pyramid 
with  basal  edge  of  two  feet  and  slant  height  of  1  yd. 

28.  Find  the   total  area  of  a   regular  quadrangular  pyramid 
with  basal  edge  of  12'  and  lateral  edge  of  10'.     Find  the  vol- 
ume. 

29.  The  base  of  a  regular  pyramid  is  a  square  with  a  side 
of  6'  and  the  lateral  area  is    f    of  the  total  area,       Find  the  alti- 
tude and  the  slant  height. 

30.  What  is   the  ratio   of  the  four  parts  into  which  a  pyra- 
mid is  divided  by  parallel  planes  dividing  the  altitude  into  four 
equal  parts? 


330  SOLID  GEOMETRY 

31.  How  much  of  a  cube  is  cut  off  by  a  plane  passing  through 
the   mid-points    of    three    concurrent   edges?      What   part    is    re- 
moved when  this  is  done  for  each  set  of  three  concurrent  edges! 

32.  If  a  plane  is   passed   through   the   extremities   of  three 
concurrent  edges  of  a  cube,  prove    (1)   that  the  tetrahedron  cut 
off  is     £     of  the  cube;    (2)   that  four  such  tetrahedrons  can  be 
cut   off   of   a  cube;    (3)    that   the   remainder  of   the   cube   is   a 
regular  tetrahedron  equal  to  £   of  the  cube. 

Sue.     Take  a  as  the  edge  of  the  cube.     Determine  the 
bases  and  altitudes  of  the  figures  and  compute  the  volumes. 

33.  Use   the   rule   for   finding   the   volume    of   a   pyramid   in 
order  to  find  the  volume  of  an  8"  cube. 

34.  How  many  cubic  feet  are  removed  in  boring  a   6"  well 
180'  deep? 

35.  How  large  an  object  at  a  distance  of  20'  will  be  obscured 
by  an  inch  square  placed  2^'   from  the  eye? 

36.  How  many  inches  from  the  vertex  of  a  cone  of  revolu- 
tion 12'  high  must  a  plane  be  passed  parallel  to  the  base  in  order 
to  bisect  the  cone?    In  order  to  trisect  it,  at  what  distances  must 
the  two  planes  be  passed? 

37.  The  interior  of  a  rectangular  bin  with  edges  in  the  ratio 
of  2,  3,  4  has  a  surface  of  672  sq.  ft.     How  many  bushels  does  it 
hold? 

THE  SPHERE. 

675.  A  SPHERE.     A  solid  bounded  by  a  surface  all 
points  of  which  are  equally  distant  from  a  fixed  point 
within  is  a  sphere. 

The  fixed  point  is  the  center  of  the  sphere  and  the 
bounding  surface  is  the  surface  of  the  sphere. 

676.  RADIUS  OP  A  SPHERE.    Any  straight  line  drawn 
from  the  center  of  a  sphere  to  its  surface  is  a  radius  of 
the  sphere. 

677.  DIAMETER   OF    A    SPHERE.      Any    straight   line 
drawn  through  the  center  of  a  sphere  and  terminated 
by  the  surface  is  a  diameter  of  the  sphere. 

678.  A  LINE  TANGENT  TO  A  SPHERE.     A  line  winch 


SPHERES  331 

touches  a  sphere  at  one  and  only  one  point  is  a  tangent 
to  the  sphere. 

679.  A  PLANE  TANGENT  TO  A  SPHERE.  A  plane  which 
touches  a  sphere  at  one  and  only  one  point  is  tangent  to 
the  sphere. 


680.  POINT   OF   TANGENCY.    The   point   in   which   a 
tangent  line  or  a  tangent  plane  touches  a  sphere  is  the 
point  of  tangency  or  point  of  contact. 

Two  spheres  are  tangent  when  they  have  one  and  only  one  point 
in  common. 

PRELIMINARY  THEOREMS. 

681.  THEOREM  I.     All  radii  of  a  sphere  are  equal 
and  all  diameters  of  a  sphere  are  equal. 

682.  THEOREM  II.     Two  spheres  are  equal  if  their 
radii  are  equal. 

In  the  case  of  spheres,  equality  implies  congruence. 

683.  THEOREM.  III.     The  radii  of  two  equal  spheres 
are  equal. 

684.  THEOREM  IV.     The  revolution  of  a  semicircle 
about  its  diameter  generates  a  sphere. 

685.  THEOREM  V.    A  straight  line   which  intersects 
a  sphere  intersects  it  in  two  points. 

686.  THEOREM  VI.    A  plane  or  a  sphere  which  inter- 
sects a  sphere  intersects  it  in  a  closed  line. 


332  SOLID  GEOMETRY 

- 

/ 
PROPOSITION  IX.  ; 

687.     THEOREM.     Every  section  of  a  sphere  by 
a  plane  is  a  circle. 


Given  the  plane  N  intersecting  the  sphere  0  in  the 
closed  line  ABD. 

To  Prove  ABD  a  circle. 

Proof.  SUG.  1.  Drop  a  1  to  plane  N  from  the  cen- 
ter Of  meeting  plane  N  in  C.  Let  A  and  B  be 
any  two  points  in  the  perimeter  of  the  section. 
Draw  AC  and  EC. 

2.  Compare  AC  and  BC. 

3.  -'-ABD  is  a  circle.    Why? 
Therefore— 

688.  A  GREAT  CIRCLE  OF  A  SPHERE.     A  circle  of  a 

sphere  which  contains  the  center  of  the  sphere  is  a  great 
circle  of  the  sphere. 

689.  A  SMALL  CIRCLE  OP  A  SPHERE.     A  circle  of  a 
sphere  which  does  not  contain  the  center  is  a  small  cir< 
of  the  sphere. 

690.  Axis   OP   A    CIRCLE   OP   A 
SPHERE.    The  diameter  of  a  sphere 
perpendicular  to  a  circle  of  a  sphere 
is   the   axis   of   the   circle   of   the 
sphere. 


SPHERES  333 

691.  POLES  OF  A  CIRCLE.  The  extremities  of  the  axis 
of  a  circle  are  its  poles. 

AB  is  a  great  circle,  EF  is  a  small  circle,  GH  is  the  axis  of 
Q  EF  and  also  of  Q  4-B  if  the  two  ©  are  parallel,  G  and  H  are  the 
poles. 

6C2.  COR.  I.  The  axis  of  a  circle  intersects  it  at  its 
center. 

693.  COR.  II.     Two  circles  of  a  sphere  equally  dis- 
tant from  the  center  are  equal. 

694.  COR.  III.    Of  two  circles  of  a  sphere  unequally 
distant  from  the  center,  that  one  ivhich  is  nearer  the 
center  is  the  greater. 

695.  COR.  IV.     The  center  of  the  sphere  is  the  cen- 
ter of  every  great  circle  of  the  sphere. 

696.  COR.  V.    All  great  circles  of  a  sphere  are  equal. 

697.  COR.  VI.     Two  great  circles  of  the  same  sphere 
intersect  in  a  common  diameter. 

698.  COR.  VII.    Two  great  circles  of  the  same  sphere 
bisect   each  other,  the  sphere,   and   the   surface   of   the 
sphere. 

699.  COR.  VIII.    Any  three  points  on  the' surf  ace  of 
a  sphere  determine  a  circle  of  the  sphere. 

SUG.     How  is  a  plane  determined? 
^  700.     COR.  IX.     Through  two  points  on  the  surface 
of  a  sphere,  not  the  extremities  of  a  diameter,  one  and 

7  7  7  7  st>J    /»^v^"f£^-*-»T          0£*SVf         xC>-< 

only  one  great  circle  can  be  drawn. 

701.  DISTANCE  ON  A  SPHERE.  The  distance  between 
two  points  on  the  surface  of  a  sphere  is  the  shorter  arc 
of  the  great  circle  joining  them. 

1.  A  plane  embracing  the  axis  of  a  circle  is  perpendicular  to 
the  circle.  If  the  arc  of  a  circle  on  a  sphere  is  bisected  by  a  plane 
embracing  the  axis,  every  point  in  the  plane  is  equidistant  from  the 
extremities  of  the  arc. 


334  SOLID  GEOMETRY 

i 

PROPOSITION  X. 

702.     THEOREM.     All   points   on   a   circle   of  a 
sphere  are  equally  distant  from  each  of  its  poles. 


Given  ABCD,  a  O  of  sphere  0,  P  and  P'  being  its 
poles,  C  and  D  any  two  points  on  the  O,  PC  and  PD 
great  circle  arcs. 

To  Prove  arc  PC^arcPD  and  arc  P'C  =  arc  P'D. 
Proof.     SUG.  1.     P  and  C  determine  a  great  circle 
which  also  passes  through  P'.    The  same  is  true 
ofPandD.    Why? 

2.  Draw  CN  and  DN,  radii  of  O  ACD. 

3.  Compare  the  chords  PC  and  PD. 

4.  Compare    the    arcs    PC    and    PD. 

5.  Compare    arcs    PCP'    and    PDP' : 
arcs  P'C  and  P'D. 

Therefore— 

703.  POLAR  DISTANCE.     The  distance  011  the  surface 
of  a  sphere  from  a  circle  to  its  nearer  pole  is  the  polar 
distance  of  the  circle. 

The  polar  distance  of  a  circle  is  less  than  a  great  semi- 
circle. 

704.  COR.  I.     The  polar  distance  of  a  great  circle  is 
a  quadrant,  i.  e.  an  arc  of  ninety  degrees. 

SUG.     What  angle  at  the  center  of  the  splinv 
subtends  the  polar  distance  of  a  great  circle? 


SPHERES  335 

PROPOSITION  XI. 

705.  THEOREM.  A  point  which  is  at  the  dis- 
tance of  a  quadrant  from  each  of  two  points  on 
the  surface  of  a  sphere,  not  the  extremities  of  a 
diameter,  is  a  pole  of  the  great  circle  embracing 
those  points. 


Given  two  points  E  and  F  on  sphere  0,  E  and  F  not 
the  extremities  of  a  diameter,  and  a  third  point  P  such 
that  arc  PE=:arePF  =  a,  quadrant. 

To  Prove  P  is  a  pole  of  the  great  circle  EF. 
Proof.     SUG.  1.     E  and  F  determine  a  great  circle. 
Why? 

2.     Join  E  and  F  to  0,  the  center  of  O 
EF,  and  P  to  0. 

3.  What    are    the    angles    POE    and 
POF1    Why? 

4.  What  is  PO  with  regard  to  O  EF*>. 
What  is  P?    Why? 

Therefore— 

NOTE— By  the  truth  of  Prop,  Til  arcs  of  great  circles 
can  be  drawn  on  the  surface  of  a  sphere  in  a  manner 
similar  to  that  by  which  arcs  can  be  drawn  on  a  plane 
with  given  radii  and  centers.  To  do  this  a  point  is  lo- 
cated at  a  quadrant's  distance  from  the  two  points 
through  which  the  great  circle  is  to  pass.  If  the  divi- 
ders be  now  opened  so  as  to  span  the  chord  subtending 


:]:;<;  SOLID  GEOMETRY 

( 

a  great  circle  quadrant  and  one  point  placed  on  the 
fixed  point  as  a  pole,  the  free  end  can  be  made  to  de- 
scribe the  desired  great  circle.  The  means  for  deter- 
mining the  quadrant  and  its  chord  will  be  found  in  §706. 

PKOPOSITION  XII. 

706.     PROBLEM.     Given   a    material   sphere,    to 
find  its  radius  or  its  diameter. 


ir 
SUG.  1.     Take  any  point  P  on  the  surface  of 

the  sphere  as  a  pole  and  with  the  dividers  de- 
scribe a  circle  C. 

2.  Take  any  three  points  A,  B,  D  on 
this  circle  and  by  means  of  the  dividers  construct 
a  AA'B'D'  congruent  to  A  ABD. 

3.  Determine  the  center  C'  of  the  cir- 
cle circumscribed  about  A  A'B'D' . 

4.  Draw  EF  equal  to  the  radius  C' A' 
and  through  E  draw  an  unlimited  straight  line 
LEF. 

5.  From  F  lay  off  FH  =  chord  PB  and 
at  F   erect   FG _L  FH,    G   being    its   intersection 
with  EH. 

6.  Prove  that  GH  is  the  diameter  of 
the  sphere  and  find  the  radius. 

707.    COR.     Given  a  diameter  of  a  sphere,  a  </uadr<int 
of  the  sphere  can  be  determined. 
Proof  left  to  the  pupil. 


SPHERES  337 

1.  Construct  on  a  plane  a  circle  equal  to  a  great  circle  of  a 
given  sphere. 

2.  Determine  the  diameter  of  a  base  ball,  a  croquet  baJl,  or 
some  other  spherical  object. 

3.  Construct  on  a  plane  a  circle  equal  to  a  given  small  cir- 
cle on  a  sphere. 

4.  To  draw  a  great  circle  that  shall  bisect  an  arc  of  a  circle 
SUG.     Locate  two  points  equidistant  from  the  extremities. 

o.         Describe   a  great   circle  through   two   points   on  the   sur- 
face of  a  sphere. 

6.  Construct  a  small  circle  on  the  surface  of  a  sphere  through 
three  given  points. 

7,  Compare  the  polar  distances  of  equal  circles  on  the  same 
sphere. 

PEOPOSITION  XIII. 

708.  THEOREM.  The  shorter  of  the  two  great 
circle  arcs  joining  two  points  on  the  surface  of  a 
sphere  is  less  than  any  other  line  on  the  sphere 
which  joins  the  two  points. 


Given  two  points  A  and  B  on  the  surface  of  the  sphere, 
AB  the  great  circle  arc  joining  them,  and  ADB  any 
other  line  on  the  surface  from  A  to  B. 

To  Prove  that  arc  AB  is  shorter  than  line  ADB. 

Proof.  SUG.  1.  Take  C  any  point  on  the  arc  AB  and 
with  A  and  B  as  poles  describe  circles  with  AC 
and  BC  as  their  respective  polar  distances.  Let 
M  l)e  any  point  on  the  first  of  these  circles  except 
point  (\  Join  M  to  A  and  to  B  by  great  arcs. 
Then  AM  +  MR  >  AB.  Why  ?  AC  =  AM  and 
hence  BM>BC.  Consequently  M  does  not  lie 


338  SOLID  GEOMETRY 

on  the  second  circle  and  the  two  circles  have  but 
the  one  point  C  in  common. 

2.  Let  D  and  E  be  the  points  in  which 
the  two  circles  meet  the  line  ADB.    As  A  is  the 
pole  of  the  O  CD,  a  line  may  be  drawn  from  A  to 
C  congruent  to  the  line  AD  and  one  can  be  drawn 
for  a  like  reason  from  B  to  C  congruent  to  line 
BE.    That  is,  a  line  can  be  constructed  from  A  to 
B  through  C  which  is  shorter  than  the  given  line 
ADB  by  the  line  DE.    Thus,  no  matter  what  line 
be  drawn  from  A  to  B  other  than  arc  AB,  C,  and 
hence  every  point  of  arc  AB}  lies  in  a  line  still 
shorter.     Therefore  every  point  of  arc  AB  lies  in 
the  shortest  line  from  A  to  B. 

3.  If  now  D  be  any  point  not  on  arc 
AB  it  cannot  lie  on  the  shortest  line  from  A  to  B. 
This  is  seen  if  the  circle  about  A  as  pole  and  with 
radius  AD  be  drawn  determining  a  point  C  by  its 
intersection  with  arc  AB.     For  then,  as  above, 
there  can  be  drawn  another  line  from  A  to  B 
through  C  which  is  shorter  than  any  line  which 
can  be  drawn  through  D. 

4.  Hence  AB  must  be  the  shortest  line. 
Therefore— 

709.  Inasmuch  as  the  great  circle  arc  is  the  shortest 
line  between  two  points  on  the  surface  of  a  sphere,  dis- 
tances on  a  sphere  are  measured  along  great  circle  arcs. 
It  will  be  natural  then  to  expect  on  the  surface  of 
a  sphere  a  surface  or  spherical  geometry  corresponding 
to  plane  geometry  with  figures  formed  from  great  circle 
arcs  instead  of  straight  lines.  This  will  appear  in  the 
sequel. 


SPHERES  339 

PROPOSITION  XIV. 

710.     THEOREM.     A   plane  perpendicular   to   a 
radius  at  its  extremity  is  tangent  to  the  sphere. 


Given  sphere  0,  radius  OG  and  plane  N  l  OG  at  G 
To  Prove  plane  N  a  tangent  plane  to  the  sphere. 
Proof.     Sue.  1.     Let  H   be   any   point    in   plane   N 
other  than  G  and  draw  OH. 

2.  Where  does  H  lie  with  respect  to 
the  sphere?    Why? 

3.  Complete  the  demonstration. 
Therefore— 

711.  COR.  I.    A  plane  tangent  to  a  sphere  is  perpen- 
dicular to  the  radius  at  the  point  of  contact. 

712.  COR.  II.     Any  straight  line  through  the  point 
of  tangency  and  in  the  tangent  plane  is  tangent  to  the 
sphere. 

713.  COR.  III.     Any  straight  line  perpendicular  to 
a  radius  at  its  extremity  is  tangent  to  the  sphere. 

NOTE.  Of  all  figures  in  plane  geometry  the  circle  most  closely 
resembles  the  sphere  in  its  characteristics.  It  is  interesting  to  note 
the  resemblance  of  certain  propositions  and  their  proofs  in  plane 
and  in  solid  geometry.  The  change  of  two  words  in  Prop.XIV 
makes  the  theorem  one  of  plane  geometry.  Compare  the  demon- 
strations of  the  two  theorems. 

Such  comparisons  of  plane  and  solid  geometry  theorems  make 
an  excellent  review  of  much  of  the  subject  and  will  also  in  many 


340  SOLID  GEOMETRY 

instances  serve  to  make  more  clear  the  meaning  and  the  demon- 
strations of  solid  geometry  theorems. 

1.  Construct  a  plane  tangent  to  a  sphere  at  a  given  point. 

2.  What  is  the  locus  of  a  point  at  a  given  distance  from  a 
given  point  and  also  equidistant  from  two  other  given  points.     Is 
the  problem  always  possible?     Is  the  locus  ever  a  single  point? 

PKOPOSITION  XV. 

714.     THEOREM.     The  intersection   of  the   sur- 
faces of  two  spheres  is  a  circle. 


Given  two  spheres  0  and  0'  intersecting  in  a  closed 
line,  two  points  of  which  are  A  and  B. 

To  Prove  line  AB  is  a  circle. 

Proof.  SUG.  1.  The  plane  AOO'  intersects  the  two 
spheres  in  two  great  circles  intersecting  each 
other  in  the  points  A  and  C.  Likewise  the  plane 
BOO'  intersects  the  two  spheres  in  two  great  cir- 
cles which  intersect  each  other  in  the  points  B 
and  D.  Why  ? 

2.  Chords  AC  and  BD  intersect  00' 
in  the  same  point  P.    Why  ? 

3.  Compare     A  AOP     with      A  BOP. 
Auth.     AP  with  BP. 

4.  AP  and  BP  lie  in  the  plane  1  00' 
at  P.   Why  ?   Hence  the  closed  line  AB  is  a  circle. 

Therefore— 

715.  COR.  I.  The  line  joining  the  centers  of  two  in- 
tersecting spheres  is  perpendicular  to  the  plane  of  their 
intersection  and  passes  through  the  center  of  the  circle 
of  intersection. 


SPHERES  341 

COB.  II.  The  plane  of  the  intersection  of  two 
equal  spheres  is  the  perpendicular  bisector  of  their  line 
of  centers. 

1.  What  is  the  locus  of  points  at  given  distances  from  two 
given  points? 

Discuss  all  possibilities. 

2.  Find  a  point  X  which  is  at  given  distances  from  two  given 
points  and  equally  distant  from  two  other  given  points.     Discuss 
all  possibilities,  showing  that  there  are  two,  one,  or  no  solutions. 

3.  Find  a  point  X  equidistant  from  two  parallel  lines,  from  two 
intersecting  lines,  and  a  given  distance  from  a  given  point.    What 
is  the  greatest  number  of  solutions?     What  is  the  least  number! 
Under  what  conditions  is  the  problem  impossible? 

716.  ANGLE  OF  Two  ARCS.     The  angle  between  the 
tangents  of  two  arcs  at  their  point  of  intersection  is  the 
angle  of  the  arcs. 

717.  SPHERICAL  ANGLE.   The  angle 
formed  by  the  arcs  of  two  great  cir- 
cles is  a  spherical  angle. 

PA  and  PB  are  two  great  circles  and 
a  and  6  are  tangent  to  the  great  circles  re- 
spectively at  P.  The  spherical  angle  APB 
equals  the  plane  angle  aPl. 

718.  COR.    A  spherical  angle  equals  the  dihedral  an- 
gle formed  by  the  planes  of  the  great  circles  forming  the 
spherical  angle. 

SUG.  The  tangents  to  the  great  circles  are  in 
the  respective  planes  of  the  circles  and  perpen- 
dicular to  the  edge  of  the  dihedral  at  the  same 
point.  Why? 

719.  A  CIRCUMSCRIBED  SPHERE.     When  all  the  ver- 
tices of  a  polyhedron  lie  in  the  surface  of  a  sphere,  the 
sphere  is  circumscribed  about  the  polyhedron  and  the 
polyhedron  is  inscribed  in  the  sphere. 


342  SOLID  GEOMETRY 

720.  AN  INSCRIBED  SPHERE.     When  the  faces  of  a 
polyhedron  are  all  tangent  to  a  sphere,  the  sphere  is 
inscribed  in  the  polyhedron  and  the  polyhedron  is  cir- 
cumscribed about  the  sphere. 

PROPOSITION  XVI. 

721.  THEOREM.     One  and  only  one  sphere  can 
be  circumscribed  about  any  tetrahedron. 


c 
Given  a  tetrahedron  ABCD. 

To  Prove  that  one  and  only  one  sphere  can  be  circum- 
scribed about  ABCD,  i.  e.  to  prove  that  there  is  one  and 
only  one  point  X  equidistant  from  A,  B}  C,  and  D. 

Proof.  SUG.  1.  What  is  the  locus  of  points  equi- 
distant from  A  and  #?  From  B  and  (7?  Auth. 

2.  Show  that  the  two  loci  just  found 
must  intersect.    What  is  then  the  locus  of  points 
equidistant  from  A}  B}  and  Cf 

3.  What  is  the  locus  of  points  equi- 
distant from  C  and  D  ? 

4.  Show  that  the  loci   in  Sug.   2  and 
Sug.  3  intersect. 

5.  Show  that   this  intersection  is  the 
required  point  X. 

6.  Show    that    only    one    such      point 
exists. 

Therefore — 


SPHERICAL  POLYGONS  343 

PROPOSITION  XVII. 

722.  THEOREM.     One  and  only  one  sphere  can 
be  inscribed  in  a  tetrahedron. 

Given  a  tetrahedron  ABCD.    §  721. 
To  Prove  that  one  and  only  one  sphere  can  be  in- 
scribed in  ABCD. 

Proof.     SUG.  1.     What  is  the  locus  of  points  equi- 
distant f rom  ABC  and  ABD  f  ABC  and  AC D  f  Auth. 
2.     Complete  the  demonstration  on  the 
outline  of  the  demonstration  §  721. 
Therefore— 

1.        All  lines   tangent   to   a   sphere   from   the  same   point   are 
equal  and  touch  the  sphere  in  a  circle  of  the  sphere. 

SUG.     Connect  the   center  of  the   sphere  with   the   given 
point  and  with  two  or  more  points  of  contact. 

PEOPOSITION  XVIII. 

723.  THEOREM.   A  spherical  angle  is  measured 
by  the  arc  of  a  great  circle  described  from  the 
vertex  of  the  angle  as  a  pole  and  intercepted  be- 
tween the  sides  of  the  angle. 


Given  two  great  circle  arcs  PA  and  PB  forming  a 
spherical  angle  at  P,  AB  being  an  arc  of  that  great 
circle  of  which  the  pole  is  P  intercepted  by  AP  and  BP 

To  Prove  that  spherical  angle  APB  is  measured  by 
arc  AB. 

Proof.  SUG.  1.  OA  and  OB  are  radii  of  the  great 
circle  ABC  and  lie  respectively  in  the  planes  of 


344  SOLID  GEOMETRY 

the  great  circles  PA  and  PB.    What  relation  do 
they  bear  to  POf    Auth. 

2.  "What  relation  does   Z  AOB  hear  to 
the  dihedral  angle  A—PO  —  Bf    Auth. 

3.  What  relation  does  arc  AB  bear  to 
£AOB?    ToZAP£?-    Auth. 

Therefore— 

724.  SPHERICAL  POLYGON.  A  portion  of  the  surface 
of  a  sphere  bounded  by  arcs  of  great  circles  is  a  spheri- 
cal polygon.  The  bounding  arcs  are  the  sides  of  the 
polygon,  the  intersections  of  the  sides  are  the  vertices, 
of  the  polygon,  and  the  spherical  angles  are  the  angles 
of  the  polygon. 

The  planes  of  the  sides  of  a  spherical  polygon  form  a 
polyhedral  angle  with  vertex  at  the  center  of  the  sphere. 
The  sides  of  a  spherical  polygon  are  great  circle  arcs 
and  may  be  expressed  in  degrees  of  arc.  They  measure 
the  corresponding  face  angles  of  the  polyhedral.  The 
spherical  angles  of  a  spherical  polygon  measure  the  dihe- 
drals of  the  polyhedral.  A  diagonal  of  a  spherical  poly- 
gon is  a  great  circle  arc  joining  any  two  non-adjacent 
vertices. 

A  diagonal  of  a  spherical  polygon  is  a  great  circle  arc 
joining  any  two  non-adjacent  vertices. 


ABCD  is  a  spherical  polygon,  AB  is  a  side,  A  is  a  vertex, 
0  —  ABCD  is  the  subtended  polyhedral  angle,  AB  measures 
L  AOB,  L  A  measures  dihedral  OA  and  AC  is  a  diagonal. 


SPHERICAL  POLYGONS  345 

725.  SPHERICAL  POLYGONS  classified.    Spherical  poly- 
gons, like  plane  polygons,  are  classified  as  triangular, 
quadrangular,  etc.,  according  to  the  number  of  angles  or 
sides.     They  may  be  right-angled,  isosceles,  equilateral, 
etc.,  as  are  plane  triangles.     They  may  be  congruent  in 
that  they  may  be  applied  to  each  other  as  are  congruent 
plane  triangles  for  two  great  circle  arcs  coincide  if  two 
points  are  common  to  them,  just  as  is  true  of  straight 
lines.    They  may  be  equal  in  area  without  being  congru- 
ent.    In  congruent  spherical  polygons  the  homologous 
parts  are  respectively  equal.    Equal  spherical  angles  may 
be  made  to  coincide. 

726.  CONVEX   SPHERICAL   POLYGON.     A 
spherical  polygon  is  convex  when  none  of 
its  sides  if  extended  will  cut  the  polygon. 

A  BCD  is  convex.  T" 

727.  CONCAVE  SPHERICAL  POLYGONS.  A  spherical  poly- 
gon which  is  not  convex  is  concave  or  reentrant.    EFGH 
is  concave. 

728.  SYMMETRICAL  SPHERICAL  TRIANGLES.  Two  spher- 
ical triangles  such  that  the  subtended  trihedral   angles 
are    symmetrical   spherical  trihedrals.  By  508  it  is  seen 
that  the  homologous  parts  of  two   symmetrical  spherical 
trihedrals  are  equal  but  arranged  in  reverse  order. 

Symmetrical  triangles  exist  in  plane  geometry  but  in 
that  case  either  of  them  could  be  removed  from  the  plane 
and  turned  over,  thus  reversing  the  order  of  its  parts, 
and  making  the  two  triangles  congruent.  The  distinc- 
tion of  congruent  and  symmetric  was  then  unnecessary. 
On  account  of  the  curvature  of  the  surface  of  a  sphere, 
no  portion  of  it  can  be  turned  over  and  applied  to  any 
other  portion.  A  distinction  between  congruent  and 
symmetrical  is  in  this  case  necessary. 


346 


SOLID  GEOMETRY 


729.  OPPOSITE  OR  VERTICAL  SPHER- 
ICAL POLYGONS.     The  polygons  inter- 
cepted, or  subtended,  on  the  surface 
of  a  sphere  by  two  opposite  or  verti- 
cal polyhedrals  with  their  vertex  at 
the  center  of  the  sphere  are  opposite 
spherical  polygons. 

ABC  and  A'B'C'  are  opposite  spherical  polygons. 

730.  CoR.I.  Opposite  spherical  polygons  are  symmetrical. 

731.  COR.  II.     Three  planes  not  having  a  common 
line  of  intersection  and  all  embracing  the  center  of  a 
sphere  intersect  the  surface  in  two  symmetrical  spher- 
ical triangles. 

PROPOSITION  XIX. 

732.  THEOREM.     Two     isosceles     symmetrical 
triangles  are  congruent. 


Given  two  symmetrical  spherical  triangles  ABC  and 
A'B'C'  isosceles  with  AB  —  BC  audA'B'  =B'C'. 

To  Prove  ABC  =  A'B'C'. 

Proof.  SUG.  1.  Compare  the  face  angles  of  the  sub- 
tended trihedrals  which  are  measured  by  AB  and 
BC;  by  A 'B'  and  B' C' .  What  kind  of  trihe- 
drals are  these? 

2.  Compare  the  two  trihedrals. 

3.  Put  the  two  trihedrals  in  coincidence 
and  compare  the  spherical  triangles. 

Therefore — 


SPHERES 


347 


1.  Demonstrate  proposition  §732  by  superposition 

733.  POLAR  OF  A  TRIANGLE.    In 
the  spherical  triangle  ABC,  let  A' 
be  that  one  of  the  two  poles  of  are 
BC  which  lies  on  the  same  side  of 
BC  as  does  the  vertex  A.     In  the 
same  manner  define  B'  and  C"  re- 
spective poles  of  arcs  AB  and  CA. 

The  triangle  A'E'C'  is  the  polar  triangle  of  ABC. 

It  is  to  be  noted  that  the  sides  of  ABC  if  extended 
will  form  eight  spherical  triangles  including  ABC.  Of 
these  but  one  answers  the  conditions  of  the  definition  ol* 
the  polar. 

2.  Draw  on  a  spherical  blackboard  or  other  sphere  eight  such 
spherical  triangles  and  distinguish  a  triangle  and  its  polar. 

PROPOSITION  XX.  . 

734.  THEOREM/    //   one   spherical   triangle   is 
the  polar  of  a  second,  then  the  second  triangle  is 
the  polar  of  the  first. 


Given  AA'B'C'  the  polar  of  A  ABC. 

To  Prove  A  ABC  the  polar  of  A  A'B'C'. 

Proof.     SUG.  1.     "What  must  be  proved  concerning  A, 

B,  C  in  order  to  show  that  A  ABC  is  the  polar  of 

A  A'B'C"? 

2.     By  §  705  prove  that  A  is  a  pole  of 

B'C',  that  B  is  a  pole  of  C'A',  and  that  C  is  a 

pole  of  A'B'. 


348  SOLID  GEOMETRY 

3.  Suppose  that  A  is  not  on  the  same 
side  of  B' C'  as  is  A'  and  draw  the  great  circle 
AA' .  Suppose  it  to  meet  BC  in  X  and  B'C'  in 
X' .  Then  both  A' AX  and  AX'  are  quadrants. 
Why?  But  as  AXf  <A'AX,  the  supposition  is 
false  and  A  and  A'  are  on  the  same  side  of  B'C' . 
Therefore— 

735.  POLAR  TRIANGLES.    Two  spherical  triangles  such 
that  each  is  the  polar  of  the  other  are  polar  triangles. 

PEOPOSITION  XXI. 

736.  THEOREM.     In   two  polar  triangles  each 
angle  of  one  is  measured  by  the  supplement  of 
the  side  opposite  it  in  the  other. 

Given  two  polar  triangles  ABC  and  A' B'C',  with 
corresponding  sides  a, 
b,  c  and  a',  &',  c'  opp- 
osite the  respective 
angles  A,  B,  C,  A', 
B',  C/. 

To  Prove  Z  .4=180° 
—  a',  etc. 

Proof.  SUG.  1.  By 
what  is  a  spherical 
angle  measured? 

2.  Extend  the  sides  of  L  A  to  meet  a' 
in  D  and  E.    Which  arc  measures  L  A  ? 

3.  a'  =  B>E  +  EC'  =  B'E  +  C'D  -  DE. 

4.  How  many  degrees  in  arc  B'E?     In 
arc  C'D?    Why? 

5.  Express  arc  DE  in  terms  of  a'. 
Therefore — 


SPHERES 


349 


1,  What  is  the  locus  of  a  point  in  space  such  that  the  sum 
of  the  squares  of  its  distances  from  two  fixed  points  equals  the 
square  of  the  distance  between  the  two  fixed  points? 

PROPOSITION  XXII. 

737.  THEOREM.  Two  symmetrical  spherical 
triangles  are  equal. 


Given  two  symmetrical  spherical  triangles  ABC  and 
A'B'C'  on  the  same,  or  on  equal  spheres. 

To  Prove  AABC  =  &  A'B'C'. 

Proof.  'Sue.  1.  Let  P  and  P'  be  the  respective  poles 
of  the  small  circles  ABC  and  A'B'C',  lying  on 
the  same  hemisphere  as  the  respective  A.  Draw 
the  great  circle  arcs  PA,  PB,  PC,  P'A',  P'B',  P'C'. 

2.  Compare  the  chords  AB,  A' B' ,  etc. 
Compare  the  circles  ABC  and  A'B'C' .    Anth. 

3.  Compare    the    arcs   PA,    PB,    PC, 
P'A',  P'B',  P'C'.     Auth. 

•A.  Compare  A  PA  B  with  AP'A'B'- 
&PBC  with  AP'B'C'  •  &PCA  with  AP'C'A'. 
Auth. 

5.  If     P     lies     within     A  ABC     then 
&ABC==&PAB+&PBC+&PCA.    If  Plies  with- 
out A  ABC  then  &ABC=PAB+ AP£C-APCA. 
Similarly  for  A  A'B'C'. 

6.  Compare    A  ABC   with    &  A'B'C'. 
Therefore — 


350 


SOLID  GEOMETRY 


PROPOSITION  XXIII. 

738.  THEOREM.  Two  triangles  on  the  same 
sphere,  or  on  equal  spheres,  having  two  sides  and 
the  included  angle  of  one  equal  to  two  sides  and 
the  included  angle  of  the  other  are  either  congru  - 
ent  or  else  symmetrical  and  therefore  equal. 


c-    c 


Given  spherical  triangles  ABC  and  A'B'C'  with 
AB=A'B',  BC=B'C',  and  LB=LB'. 

To  Prove  A  ABC  =  A  A'B'C'  when  the  order  of  ar- 
rangement of  the  respective  parts  is  the  same  and 
A  ABC  symmetrical  to  A  A 'B 'C'  when  the  order  oF 
arrangement  is  different. 

Proof.      CASE  I.      SUG.       Superimpose     A  ABC    on 
A  A'B'C'.  See  §  725. 

CASE  II.  SUG.  1.     Construct  A  A"B"C"  sym- 
metrical to  A  ABC. 


2.  Compare    A  A"B"<" 
A  A'B'C'.     Case  I. 

3.  Compare       A  ABC 

§728. 


with 


with 
A  A'B'C'. 

Therefore— 

739.  EQUAL  POLYHEDRALS.  Two  polyhedrals  which 
when  placed  with  their  vertices  at  the  center  of  the 
same  sphere  intersect  on  the  surface  equal  spherical  poly- 
gons are  equal  polyhedrals,  or  ctjunl  xolid  angl<  *. 


SPHERES  351 

740.  COR.    Two  trihedrals  having  a  dihedral  and  the 
including  face  angles  of  one  equal  respectively  to  a  di- 
hedral and  the  included  face  angles  of  the  other  are  either 
congruent  or  else  symmetrical  and  equal. 

SUQ.    Use  Prop.  §  738,  and  §  718. 

PROPOSITION  XXIV. 

741.  THEOREM.     Two  triangles  on  the  same,  or 
on  equal,  spheres  having  two  angles  and  the  in- 
cluded side  of  one  equal  to  two  angles  and  the 
included  side  respectively  of  the  other  are  either 
congruent   or    else    symmetrical    and    therefore 
equal. 


c- 

Given  A  ABC  and  A'B'C'  on  the  same  or  on  equal 
spheres,  with  L  A=  L  A' ,  LE—LB',  and  arc  AB  = 
arc  A'B'. 

To  Prove  A  ABC  and  A'B'C'  either  congruent  or 
else  symmetrical  and  equal. 

Proof.  SUG.  1.  Let  &  EFG  and  E'F'G'  be  the  re- 
spective polars  of  ABC  and  A'B'C'  with  sides 
e,  1,  9,  e',  /',  9',  opposite  A  E,  F,  G,  E',  F',  G'  re- 
spectively. The  pupil  may  construct  the  polar 
triangles. 

2.  Thene=e',  /=/',  LG=LG'. 

3.  Then  by  §  738  A  EFG  and  E'F'G' 
are  either  congruent  or  symmetrical.     Homologous 
parts  of  A  EFG  and  E'F'G'  are  then  equal. 


352  SOLID  GEOMETRY 


4.  Since  LE=LE\LV=LF  and  g=g', 
it  follows  that  a—a',  &=&',  and  ZC=ZC". 

5.  Compare  4  ABC  and  A'B'C'.  Auth. 
Therefore— 

742.  COR.     Two  trihedrals  having  two  dihedrals  and 
included  face  angle  of  one  equal  to  two  dihedrals  and 
the  included  face  angle  of  the  other  respectively  are 
either  congruent  or  symmetrical. 

PROPOSITION  XXV. 

743.  THEOREM.     Two    triangles    on   the   same 
sphere,  or  on  equal  spheres,  having  the  three  sides 
of  one  equal  respectively  to  the  three  sides  of  the 
other  are  either  congruent  or  else  symmetrical 
and  therefore  equal. 


Given  two  spherical  triangles  ABC-  and  A'B'C'  on 
the  same  or  equal  spheres  with  AB  =  A'B',  BC  =  B'C', 
CA  =  C'A'. 

To  Prove  A  ABC  and  A'B'C'  either  congruent  or 
else  symmetrical  and  equal, 

Proof.  SUG.  1.  Connect  the  vertices  with  the  re- 
spective centers  0  and  0' . 

2.  Compare  the  face  angles  of  the  re- 
spective trihedrals  0  and  0'. 

3.  Compare  the  trihedrals  0  and  0'. 

4.  Compare    the    triangles    ABC    and 
A'B'C'. 

Therefore— 


SPHERES  353 

PKOPOSITION  XXVI 

744.  THEOREM.     Two    triangles   on   the   same 
sphere,  or  on  equal  spheres,  having  the  three  an- 
gles of  one  equal  respectively  to  the  three  angles 
of  the  other  are  either  congruent  or  else  symmet- 
rical and  therefore  equal. 

Given  two  triangles  ABC  and  EFG  on  the  same  or 
equal  spheres,  with  LA=LE,  LE—LF,  ZC=Z£  with 
respective  sides  a,  1),  c,  c,  f,  g. 

The  pupil  may  construct  the  figure  from  description  in  the  text. 

To  Prove  A  ABC  and  EFG  congruent  or  else  sym- 
metrical. 

Proof.     SUG.  1.     Let  A  A 'B' C'  and  E'F'G' be  the  re 
spective   polars  of   A  ABC  and  EFG,  with   respective 
sides  a',  V,  c',  e ,  /',  g'.     Compare  a'  and  e't  I)'  and  /',  c' 
and  g'.     Auth. 

2.  Compare  A  A'B'C'  and  E'F'G'.  Auth. 

3.  Compare  1A',  £',  C"  with  AE',F',G' 
respectively.    Auth. 

4.  Compare  sides  a,  b,  c  with  e,  f,  g 
respectively.    Auth. 

5.  Compare  A  ABC  and  EFG.     Auth. 
Therefore— 

745.  COR.    Two  trihedrals  having  the  three  dihedrals 
of  one  equal  to  the  three  dihedrals  of  the  other  respec- 
tively are  either  congruent  or  symmetrical  and  there- 
fore equal. 

1.  A  straight  line  tangent  to  a  circle  of  a  sphere  lies  in  the 
plane  which  is  tangent  to  the  sphere  at  the  point  of  contact. 

2.  What   is   the   locus  of   a   line    tangent    to    a  sphere   at   a 
given  point? 

3.  By  planes  parallel  to  the  base  divide  a  pyramid  into  two 
equal  parts. 


354  SOLID  GEOMETRY 

4.  If  two  angles  of  a  spherical  triangle  are  equal,  the  tri- 
angle is  isosceles. 

Sue.     Construct  the  polar  of  the  given  triangle. 

5.  The  arc  of  a  great  circle  drawn  from  the  vertex  of  an 
isosceles  triangle  to  the  middle  of  the  base  is  perpendicular  to  the 
base  and  bisects  the  vertex  angle. 

6.  Determine  a  point  X  at  a  given  distance   from  a  fixed 
point,  equidistant  from  two  parallel  planes,  and  equidistant  from 
two  given  points. 

7.  Prove  Prop.  §  741  by  superposition. 

8.  If  one  circle  of  a  sphere  passes  through  the  poles  of  a 
second  circle  of  a  sphere,  the  planes   of  the  two  circles  are  per- 
pendicular to  each  other. 

9.  Find  a  point  X  at  a  distance  m  from  one  point,  a  dis- 
tance n  from  a  second  point,  and  a  distance  p  from  a  third  point. 
V\rhen  is  there  no  solution?     When  is  there  one?    When  two?     Is 
there  any  other  possibility? 

10.  The    angles    opposite    the    equal    sides    of    an    isosceles 
spherical  triangle  are  equal. 

11.  In  a  given  plane,  find  a  point  which  is  equidistant  from 
the  vertices  of  a  triangle  which  is  in  another  plane.     Use  locus. 

PKOPOSITION  XXVII. 

746.  THEOREM.     I.    Each  side  of  a  spherical 

triangle  is  less  than  the  sum  of  the  other  two. 

SUG.     Construct  the  subtended  trihedral  angle 
at  the  center  of  the  sphere  and  use  §  509. 

PKOPOSITION  XXVIII. 

747.  THEOREM.     The  sum  of  the  sides  of  a  con- 
rex  spherical  polygon  is  less  than  360°  of  arc. 

SUG.     Construct  the  subtended  polyhedral  and 
use  §-510. 

748.  COR.    The  sum  of  the  sides  of  a  convex  spherical 
polygon  iff  less  than  a  great  circle. 


SPHERES  355 

749.  COR.    No  side  or  diagonal  of  a  convex  spherical 
polygon  is  as  great  as  180°  of  arc. 

PROPOSITION  XXIX. 

750.  THEOREM.     The  sum  of  the  angles  of  a 
spherical  triangle  is  greater  than  180°  and  less 
than  540°. 


c- 

Given  A  ABC. 

To  Prove    ZA+ZJ2  +  ZO  180°   and  Z  A  +  Z£  + 
Z  C  <  540°. 
Proof.     SUG.  1.     Let    kA'B'C'     be    the    polar    of 

&ABC.       Then    A  =  180°—  a',    B  =  l80°—l', 

C=l80°—c'.    Why? 

2.  Then         A  +  B  +  C  =  540°  - 
(a'  +  fc'+c')     Why? 

3.  a'  +  &'  +  c'  <  360°.     Why? 

4.  Hence  ZA  +  LE  +  L  C  greater  than 
180°andZA-fZ£  +  ZC'<  540°. 

Why? 
Therefore— 

1.  The  angles  of  a  spherical  triangle  are  70°,  96°,  and  84°. 
How  many  degrees  in  the  respective  sides  of  the  polar  triangle? 

2.  The  sides  of  a  triangle  are  90°,  65°,  and  125°.    How  many 
degrees  in  the  respective  angles  of  the  polar  triangle? 

751.  SPHERICAL  EXCESS.  A  spherical  triangle,  unlike 
a  plane  triangle,  may  have  two  or  three  right  angles,  or 
two  or  three  obtuse  angles.  The  number  of  degrees  in 
the  angles  of  a  spherical  triangle  in  excess  of  two  right 
angles  is  the  spherical  excess  of  the  triangle. 


356  SOLID  GEOMETRY 

i 

752.  COR.     The  spherical  excess  of  a  spherical  tri- 
angle is  less  than  four  right  angles. 

753.  LUNE.    A  portion  of  the  surface  of  a 
sphere  included  between  two  semicircles  is  a 
lune.     The  angle  between  the  great  circles  is 
the  angle  of  the  lune. 

ABCDA   is  a  lune,  A   and   C  are  its  angles    and 
ABC,   ADC  are  its  edges. 

PRELIMINARY  THEOREMS. 

754.  THEOREM  I.    The  angle  of  a  lune  is  equal  to  the 
dihedral  angle  of  the  planes  of  the  great  circles  forming 
the  lune. 

755.  THEOREM  II.    The  angle    of  a  lune  is  measured 
by  the  arc  of  a  great  circle  described  from  either  vertex 
as  a  pole  and  intercepted  between  the  sides  of  the  lune. 

756.  THEOREM  III.    Two  lunes  on  the  same  sphere,  or 
on  equal  spheres,  are  equal  if  their  angles  are  equal. 

1.  What  is  the  locus  of  a  point  which  is  at  a  given  distance, 
a,  from  a  given  plane,   and  at  a   given   distance  greater  than  a 
from  a  given  point  in  the  given  plane? 

757.  TRI-RECTANGULAR  TRIANGLES.    The  eight  spheri- 
cal triangles  into  which  a  spherical  surface  is  divided  by 
three  great  circles  the  planes  of  which  are  perpendicular 
to  one  another  are  tri-rectangidar  triangles,  i.  e.,  each 
contains  three  right  angles. 

758.  SPHERICAL  DEGREE.    If  a  tri-rectangular  triangle 
be  divided  into  ninety  equal  parts,  one  of  these  parts  is 
a  degree  of  surface  or  spherical  degree. 

2.  One    three    hundred    sixtieth    part   of    the    surface    of    a 
hemisphere,   or  one  seven  hundred  twentieth  part  of  the  surface 
of  a  sphere/  is  a  spherical  degree. 


AREA  OF  SPHERES  357 

PEOPOSITION  XXX. 

759.  THEOREM.  The  surface  of  a  lune  is  to 
the  surface  of  the  sphere  as  the  angle  of  the  lune 
is  to  four  right  angles. 


Given  a  sphere  0  with  surface  S  and  a  lune  with  an- 
gle BAG  and  surface  S' . 

S'      LBAC 
To  Prove  —  =  —      — . 

8      4  rt.  4 

Proof.     CASE  I.     L  B AC   commensurable  with   4  rt. 
A.  or  360°. 

SUG.  1.  From  A  as  a  pole  draw  the  great  cir- 
cle BC.  Since  the  angles  at  A  are  measured  by 
the  arcs  which  they  intercept  on  this  great  circle, 
arc  BC  and  the  great  circle  BC  are  commensur- 
able. Divide  the  arc  BC  and  the  circle  BC  by  a 
common  unit  of  measure  and  through  each  point 
of  division  and  A  pass  a  great  circle.  Compare 
the  small  lunes  into  which  lune  BAG  and  the 
spherical  surface  are  divided. 

2.  Using  one  of  these  lunes  as  a  unit 

compare  the  ratio  —  with  the  ratio  - 

8  OBC 

3.  Using  the  angle  of  the  unit  lune  as  a 

/  Ft  4  C1 

unit,  compare  the  ratio  -'- — ^-  with  the  ratio 


358  SOLID  GEOMETRY 


OBC 


LBAC 


4.     Compare  the  ratio    :-  with  the  ratio 


360° 

CASE  ii.   L  BAC  not  commensurable  with  4 
rt.  £  or  360°. 

SUG.     Proceed  as  in  428. 

760.  COR.  I.     The  number  of  spherical  degrees  in  a 
lune  is  double  the  number  of  angular  degrees  in  its  angle. 

SUG.     Let  S  denote  the  number  of  spherical  de- 
grees in  the  lune  and  A  the  number  of  angular 

S        A 

degrees  in  the  angle.     Then  whence 

720      360 

#=2,1. 

761.  COR.  II.    The  surface  of  a  lune  equals  the  prod- 
uct of  a.  tri-rectangular  triangle  by  twice  the  number  of 
right  angles  in  the  angle  of  the  lune. 

SUG.     Denote   the   tri-rectangular   triangle   by 
T  and  the  number  of  right  angles  in  the  angle  of 

the  lune  by  A.     Then  - 

4  rt.  A.       o± 

•'•S  =  2AX  T. 

1.  The  angle  of  a  lune  is  72°  and  the  area  of  the  sphere  is 
95  sq.  in.     Find  the  area  of  the  lune. 

SUG.     By  Cor.  I,     S  =  2a  sph.  deg.  =  2  x  72  sph.  detfnv.~ 
Cor.  II,  S  =  2  A  x  T  =         ^—-  T. 

2.  The  angle  of  a  lune  is  130°   and  the  area  of  the  sphere 
is  45  sq.  yds.     Find  the  number  of  square  yards  in  the  lune. 


AREA  OF  SPHERES  359 

PROPOSITION  XXXI. 

762.  THEOREM.  The  sum  of  the  areas  of  the 
two  vertical  spherical  triangles  formed  on  a  hem- 
isphere by  two  intersecting  great  circles  equals  a 
lune  with  an  angle  equal  to  the  angle  of  the  inter- 
acting circles. 


Given  a  hemisphere  0  —  ABCD  with  two  great  cir- 
cles ACE  and  BCD  intersecting  at  C  and  forming  two 
vertical  spherical  A  ACD  and  BCE. 

To  Prove  AACD  +  A  ECB  =  lune  CAFD. 

Proof.     SUG.     Compare    A  ECB    with    its    opposite, 
&AFD. 

Therefore— 

PROPOSITION  XXXII. 

763.  THEOREM.  The  number  of  spherical  de- 
grees in  a  spherical  triangle  equals  the  number  of 
angular  degrees  in  its  spherical  excess. 


Given  A  ABC,  S  denoting  its  area  in  spherical  de- 
grees, A,  B,  and  C  denoting  the  number  of  angular  de- 


360  SOLID  GEOMETRY 

t 

grees   in   the   angles   of    A  ABC,   and   E   denoting   its 
spherical  excess. 
To  Prove  8  =  E. 

Proof.     Sue.  1.     The     surface     of     the     hemisphere 
A  —  BCC'B',  or  360  sph.  deg.,  equals 
A$  +  AJ/  +  AAT  +  &P,H,N,P  and  8,  denoting 
the  number  of  spherical  degrees  in  the  respective 
triangles. 

2.  A  8  +  A  M  =  lune  B, 

AS  +  AN  =  lune  C,  A  8  +  A  P  =  lune  A.    Why  ? 

3.  Hence     S  +  M  =  2B,     S  +  N  =  2  C, 
S  +  P  =  2  A,  .:  2  S  +  360  =  2  (A+B+C).     Why  ? 

4.  Whence  S  =  A  +  B  +  C  —  180,  i.  e. 
the    number    of    spherical    degrees    in    A  ABC 
equals  the  number  of  angular  degrees  in  the  tri- 
angle less  180°,  i.  e.  S  =  E. 

Therefore— 

764.  COR.  I.     The   ratio   of  a  spherical   triangle   to 

1? 

the  surface  of  its  sphere  is    — —    and  its  ratio  to  a  tri- 

pt 

rectangular  triangle  is  — 

8 

1.  The  area  of  a  spherical  triangle  can  be  obtained  when  the 
area  of  the  sphere  and  the  angles  of  the  triangle  are  known. 

2.  The  angles  of  a  triangle  are  96°,  72°  and  120°,  and  the 
surface  of  the  sphere  is  360  sq.  in.     How  many  sq.  in.  in  the  sur- 
face of  the  triangle? 

765.  SPHERICAL   EXCESS   OF  A   SPHERICAL  POLYGON. 
The  number  of  degrees  by  which  the  sum  of  the  angles 
of  a  spherical  polygon  of  n  sides  exceeds  (n  —  2)  180° 
is  the  spherical  excess  of  the  polygon. 


AKEA  OF  SPHERES  361 

PROPOSITION  XXXIII. 

766.  THEOKEM.  The  number  of  spherical  de- 
grees in  a  spherical  polygon  equals  the  number  of 
angular  degrees  in  its  spherical  excess. 


o 

Given  a  spherical  polygon  ABCD...,  S  denoting  its 
area  in  spherical  degrees  and  E  its  spherical  excess. 

To  Prove  S  =  E. 

Proof.  SUG.  1.  Draw  all  possible  diagonals  from 
some  one  vertex,  thus  dividing  the  polygon  in.to 
n  —  2  A.  Denote  their  areas  in  sph.  deg.  by  $15  S2, 
S3,  etc.,  and  their  respective  spherical  excesses  by 
Elt  E2,  E3,  etc. 

=  (4  of  A^-  180°)  +  (4  of  ASa  -  180°)  +  .  .  .  . 
=  4  of  ABCD  -  (n  -  2)  180°  =  E. 
4.     S  =  Sl  +  S2  +8B  H-  .  . 

Therefore— 

767.  •  A  ZONE.  A  portion  of  the  surface  of  a  sphere 
included  between  two  parallel  planes  is  a  zone.  The  cir- 
cles of  the  sphere  formed  by  the  bounding  planes  are 
the  bases  of  the  zone  and  the  distance  between  them  is 
the  altitude  of  the  zone. 


362 


SOLID  GEOMETRY 


768.  SPHERICAL  SEGMENT.    A  portion  of  a  sphere  in- 
cluded between  two  parallel  planes  is  a  spherical  seg- 
ment.    The  sections  of  the  sphere  formed  by  the  two 
planes  are  the  bases  of  the  segment  and  the  distance  be- 
tween the  planes  is  the  altitude  of  the  segment. 

In  the  figure  X  is  a  spherical  segment.  The  spherical  surface 
of  a  segment  is  a  zone.  If  a  portion  of  a  sphere  be  cut  off  by 
a  plane,  this  portion  is  a  segment  and  its  spherical 
surface  is  a  zone,  for  they  are  included  between 
the  cutting  plane  and  a  tangent  plane  parallel  to 
the  cutting  plane.  In  this  case  the  segment  and 
the  zone  have  but  one  base  each.  Find  illustra- 
tions of  zones  of  both  kinds  from  geography. 

769.  If  a  semicircle  be  revolved  about  its  diameter  as 
;.n  axis  a  spherical  surface  is  generated.    Any  arc  of  the 
semicircle  generates  a  zone. 

770.  SPHERICAL  SECTOR.    A  portion  of  a  sphere  gen- 
orated  by  the  revolution  of  a  circular  sector  about  a 
diameter  is  a  spherical  sector. 

It  is  important  to  form  mental  pictures  of  the 
different  varieties  of  spherical  sectors  and  de- 
scribe them.  For  example,  if  the  semi-circle  ADB 
is  revolved  about  the  diameter  AB,  the  circular 
sector  AOC  generates  a  spherical  sector  the  sur- 
face of  which  consists  of  a  zone  of  one  base  gen- 
erated by  the  arc  AC  and  a  convex  conical  surface 
generated  by  the  radius  OC.  The  sector  COD 
generates  a  spherical  sector  6  bounded  by  a  zone 
of  two  bases,  a  convex  conical  surface  generated 
by  OD,  and  a  concave  conical  surface  generated 
by  OC. 

Generate    many    spherical    sectors   and    describe    them, 
many  drawings  to  illustrate  spherical  sectors. 

1.  Construct  a  semi-circle  and  in  it  a  circular  sector  which, 
if  revolved,  will  generate  a  spherical  sector  having  two  concave 
conical  surfaces.  Describe  the  zone. 


AREA  OF  SPHERES 


363 


1.  Construct   a    circular    sector   which    generates    a   spherical 
sector  the  surface  of  which  consists  of  a  concave  conical  surface, 
a  plane  surface,  and  a  zone. 

2.  Is  a  hemisphere  a  spherical  sector?     Why?     A  spherical 
segment?     Why? 

3.  Describe  the  sectors  that  have  for  their  spherical  surface 
the  torrid  zone,  the  N.  frigid  zone,  and  the  S.  temperate  zone. 

PROPOSITION  XXXIV. 

771.  THEOREM.  The  area  of  the  surface  gen- 
erated by  the 'revolution  of  a  straight  line  seg- 
tnent  about  an  axis  in  its  plane  but  not  crossing 
the  axis,  is  equal  to  the  product  of  the  projection 
of  the  segment  upon  the  axis  and  the  circumfer- 
ence of  a  circle  the  radius  of  which  is  a  perpen- 
dicular erected  at  the  mid  point  of  the  segment 
and  terminated  by  the  axis. 


Ayjc 


Ar 1C 

d 
N 


—  — i D 


Given  axis  d  with  a  line  segment  AB  in  the  same 
plane  as  d,  but  not  crossing  it,  CD  the  projection  of  AB 
on  the  axis,  M  the  mid  point  of  AB,  MO  a  perpendicular 
to  AB  at  M  intersecting  the  axis  at  0  and  8  the  area  of 
the  surface  generated  by  the  revolution  of  AB  about  the 
axis. 

To  Prove  8  =  2*  X  MO  x  CD. 

Proof.  CASE  I.  AB  oblique  to  CD  not  intersect- 
ing it. 


364  SOLID  GEOMETRY 

SUG.  1.     Draw  MN 1  CD,  join  A  and  Cf  B  and 
D,  draw  AP  II  CZ>.    AC  and  £Z)  are  1  CD.    Why  ? 

2.  £  =  27r  X  MN  X  AB.     Why? 

3.  A  ABP  —  A  OMN.     Why  ? 
•'•  ABXMN=CDX  MO,    Why  ? 

4.  •'•  #  =  27r  X  MO  X  CD. 
CASE  II.     When  AB  meets  CD. 

SUG.     Follow  the  plan  of  Case  I. 
CASE  III.     When  AB  II  CD. 
Proof  left  to  the  pupil. 

Therefore— 

1.  How  many  spherical  degrees  are  there  in  a  spherical  tri- 
angle with  angles  of  200°,  140°,  and  100°? 

2.  What  part  of  the  surface  of  a  sphere  is  a  spherical  tri- 
angle with  angles  of  120°,  140°,  and  160°? 

772.  POSTULATE.  If  half  of  a  regular  polygon  be  in- 
scribed in  a  semicircle  and  the  number  of  sides  be  indefi- 
nitely increased,  and  if  the  semicircle  be  revolved  about 
its  diameter  the  surface  generated  by  the  semipolygon 
is  a  variable  which  approaches  the  sphere  generated  by 
the  semicircle  as  its  limit  and  the  apothem  of  the  poly- 
gon approaches  the  radius  of  the  sphere  as  its  limit. 

3.  A  boiler  is  4$  ft.  in  diameter  and  18  ft.  long.    It  is  pene- 
trated  by  48    3-inch  cylindrical  tubes.     How  many   gallons   does 
it  hold! 

4.  A  pyramid  with  an  altitude  of     T3ff     ft.  is  cut  into  two 
parts  of  equal  volume  by  a  plane  parallel  to  the  base.     Find  the 
distance  of  the  cutting  plane  from  the  vertex. 

5.  A  plane  parallel  to  the  base  of  a  cone  bisects  the  alti- 
tude.   Compare  the  volumes  of  the  two  parts. 

6.  A  circular  water  tank  with  a  diameter  of  .10^'  is  2i'  deep. 
How  many  barrels  will  it  hold?    (Indicate  operation  and  abbreviate 
by  cancellation.) 


AREA  OF  SPHERES  365 

PROPOSITION  XXXV. 

773.     THEOREM.     The  area  of  a  sphere  is  equal 
to  the  product  of  its  diameter  and  a  great  circle. 


B 

Given  a  sphere  0  generated  by  the  revolution  of  the 
semicircle  ACDB,  its  surface  denoted  by  S,  its  radius 
by  r. 

To  Prove  8  =  2*  r  x  2r. 

Proof.  SUG.  1.  Divide  arc  ACDB  into  equal  parts 
and  draw  the  chords  forming  the  regular  semi- 
polygon  ACDB.  Draw  the  projections  of  the 
chords  upon  the  diameter  AB  represented  AC', 

C'  V Draw  the  apo- 

them  x  of  the  polygon  to  each  chord. 

2.  The  surface  generated  by  AC  equals 
27r  X  AC'  X  x.     Why?     What  is  the  surface  gen- 
erated by  CD?    By  DB°! 

3.  If  8'   be  the  surface  generated  by 
the.  semi-polygon,  then  8'  =  2-n-  X  AB  X  x.    Why  ? 

4.  If   the   number   of   sides   be   indefi- 
nitely increased  then  S'  =:  S,  x  =  r  2?r  AB  x  =  2?r 
AB  x  >  •'•  tf  =- 2,rr  *  2r.  Why? 

Therefore— 

774.  COR.  T.     The    .surface    of   a  sphere    equals   4irra. 
or  "V/-. 

775.  COR.  II.     The  area  of  the  surface  of  a  sphere 
equals  that  of  four  great  circles. 


366  SOLID  GEOMETRY 

t 

776.  COR.  III.  The  area  of  the  surface  of  a  sphere 
equals  that  of  a  circle  with  a  radius  equal  to  the  diam- 
eter of  the  sphere. 

111.  COR.  IV.  The  surfaces  of  two  spheres  have  the 
same  ratio  as  the  squares  of  their  radii  and  the  squares 
of  their  diameters. 

4:7rl'~ 

778.  COR.  V.     The  area  of  a  spherical  degree  is  — 

779.  COR.  VI.     The  area  of  a  spherical  triangle  is 

. 
,  in  which  b  is  the  spherical  excess  of  the  triangles. 


720 

780.     COR.  VII.     2  lie  area  of  a  spherical  polygon  is 

-1      .,  2  V  J7* 

,  in  which  E  is  the  spherical  excess  of  the  polygon. 


781.  COR.  VIII.     The  area  of  a  zone  is  2-n-r  x  hf  ;n 
which  h  is  the  altitude  of  the  zone. 

SUG.     Demonstrate  according  to  the  method  of 
773,  using  an  arc  less  than  a  semicircle. 

782.  COR.  IX.    Zones  on  the  same  sphere  or  on  equal 
spheres  are  proportional  to  their  altitudes. 

783.  POSTULATE.     //  a  polyhedron  be  circumscribed 
about  a  sphere  and  the  number  of  its  sides  be  indefinitely 
increased  in  such  a  manner  that  the  areas  of  each  of  the 
faces  of  the  polyhedron  are  at  the  same  time  indefinitely 
decreased,  the  surface  of  the  polyhedron  is  a  variable 
which  approaches  the  surface  of  the  sphere  as  its  limit 
and  the  volume  of  the  polyhedron  is  a  variable  which 
approaches  the  volume  of  the  sphere  as  its  limit. 

784.  SPHERICAL  PYRAMID.     A  solid    bounded    by    a 
spherical   polygon  and  the  polyhedral  angle  which  it 
subtends  at  the  center  of  the  sphere  is  a  spherical  pyra- 
mid.   Its  vertex  is  the  center  of  the  sphere  and  its  base 
is  the  spherical  polygon, 


VOLUME  OF  A  SPHERE  367 

PKOPOSITION  XXXVI. 

785.  THEOREM.     The   volume    of  a    sphere   is 
equal  to  the  area  of  its  surface  multiplied  by  one- 
third  its  radius. 

Given  a  sphere,  denoting  its  volume  by  V,  its  surface 
by  S,  and  its  radius  by  r. 

To  Prove  V  =  $rxS. 

Proof.  SUG.  1.  Circumscribe  a  polyhedron  about 
the  sphere  and  join  each  vertex  to  the  center. 
Each  face  forms  the  base  of  a  pyramid,  the  edges 
of  which  are  the  lines  from  its  vertices  to  the 
center.  What  is  the  altitude  of  each  pyramid  1 

2.  What  is  the  volume  of  each  pyra- 
mid ?    What  is  the  volume  of  the  polyhedron  ? 

3.  Apply  the  limiting-  process  of  §783. 

Therefore— 

786.  COB.  I.   The     volume     of    a     sphere     i,s     jTrr3 
(774), r  being  the  radius ,  or  JTT  D*. 

787.  COB.  II.     The  volumes  of  tu\o  spheres  have  the 
same  ratio  as  the  cubes  of  their  radii  and  the  cubes  of 
their  diameters. 

788.  COB.  III.     The  volume  of  a  spherical  sector  is 
equal  to  the  product  of  the  area  of  its  zone  and  one-third 
the  radius  of  the  sphere,  i.  e.  $*  R2  h. 

SUG.     Base  a  demonstration  upon  the  method 
of  785. 

789.  COB.  IV.     The  volume  of  a  spherical  pyramid 
equals  the  product  of  the  area  of  its  base  and  one  third 

fvr9S 

the  radius  of  the  sphere,  ?.  c.   -       — 

720 

SUG.  1.     Make   a   demonstration  based  on  tho 
method  of  785. 


368 


SOLID  GEOMETRY 


2.     V- 


720 


PROPOSITION  XXXVII 

790.     PROBLEM.     To  find  tlie  volume  of  a  spheri- 
cal segment. 


Given  a  spherical  segment  generated  by  the  revolution 
of  arc  AC  about  the  diameter  OP,  r±  and  r2  denoting  the 
radii  of  its  respective  bases,  h  its  altitude,  r  the  radius 
of  the  sphere,  and  V  the  volume  of  the  segment. 

CASE  I.  To  find  V  in  terms  of  r\,  r2,  and  h1}  when 
both  bases  are  on  the  same  side  of  the  center, 

SUG.  1.  Let  m  represent  OD,  the  distance 
from  the  center  to  the  nearest  base,  and  let  Vi, 
V2,  Vs  denote  the  volumes  generated  by  the  tri- 
angles OAB,  OCD,  and  the  sector  CO  A  respec- 
tively. 

2.      y=F  +  F1-Fa.     Why? 


—  -J  [27rr2/*  +  (r  2  -  mTI2)  (m  +  //)  - 
(r2—  w2)w],  which  by  expanding  and  removing 
parentheses  becomes 

=  J  -IT  [3r  2  -  3  wi  2  -  3  A  ?>i  -  /*.  a  ]  // 

=Wi  [r 


m 


2—       — 


EXERCISES  369 

3.  In  this  last  expression  put  the  first 
r~  —  w2  equal  to  r22  (Auth.)  and  the  second  one 
equal  to  r-A2  +  2hm  +  h2,  for 


4.     .•-.     F=|/i[7r 
Therefore— 

The  volume  of  a  spherical  segment  equals  one-half  the 
product  of  its  altitude  and  the  sum  of  its  bases,  increased 
by  one-sixth  the  volume  of  a  sphere  with  a  diameter 
equal  to  the  altitude  of  the  segment. 

CASE  II.     When  the  center  is  between  the  bases. 

SUG.     Adapt  the  work  of  case  I  to  this  case. 
(\\SE  III.     When  the  segment  has  but  one  base. 

SUG.     Adapt  the  work  of  case  I  to  this  case  by 
letting  r,  be  zero.     What  term  will  vanish  from 
the  final  formula?     Translate  the  formula  into 
words. 
Use  3}  for  IT  in  the  following  exercise: 

1.  Given  a  sphere  20'    in  diameter.      Find   the   volume  of  a 
segment,  the  radii  of  the  bases  being  6  and  8  ft.  respectively. 

2.  Given  a  sphere  20'    in  diameter.     Find  the   volume  of  a 
segment  with  one  base  having  a  radius  of  8'. 

3.  A  4-in.   hole   is   bored   through   a   sphere    10"   in    diameter. 
How  much  of  the  volume  of  the  sphere  is  cut  away? 

4.  Find    the    volume    of   a    spherical    sector    having    a    zone 
with  an  altitude  of  10"  on  a  sphere  with  a  radius  of  20". 

5.  Find  the  volume  and  area  of  a  sphere  40"  in  diameter. 

6.  A  sphere  is  cut  by  parallel  planes  so   that  the  diameter 
is  divided  into  ten  equal  parts.     Compare  the  areas  of  the  zones 
and  also  the  volumes  of  the  spherical  sectors  the  spherical  surfaces 
of  which  are  the  respective  zones. 


370  SOLID  GEOMETRY 

7.  If  the  average  specific  gravity  of  the  earth  is  5.6,  what 
is  its  weight  in  tons? 

8.  Find  the  angles  of  an  equiangular  spherical  triangle,  the 
surface  begin  -fa  that  of  the  sphere. 

9.  The  radius  of  a  sphere  is  3  in.  and  the  area  of  a  spherical 
triangle  ABC  on  the  sphere  is  12  sq.  in.     The  angles  A   and   Tt 
are  140°  and  115°  respectively.     Find  Z  C. 

10.  The  dimensions  of  a  rectangular  parallelepiped  are  3',  4'. 
and  12'.     Find  the  length  of  a  great  circle  of  the  circumscribing 
sphere. 

11.  Assume  the  diameters  of  the  earth  and  moon  to  be  8,000 
and  2,000  miles  respectively.    What  is  the  ratio  of  their  volumes! 

12.  A  triangle  on  a  12"  globe  has  angles  of  140°,  119°,  and 
196°.    Compute  the  area. 

13.  The  angles   of   a   pentagonal  spherical  pyramid  are   47°, 
96°,  120°,  142°,  and  87°,  and  the  diameter  of  the  sphere  is  20". 
Find  the  area  of  the  base  and  the  volume  of  the  pyramid. 

14.  Prove  that  the  volume  of  a  portion  of  a  sphere  included 
by  the  planes  of  two  great  circles  and  the  intercepted  lune  is  to 
the  volume  of  the  sphere  as  the  angle  of  the  lune  is  to  4  rt.  angles. 
Such  a  solid  is  a  wedge  or  ungula.    The  lune  is  its  base. 

15.  A    cone   of   revolution   and    a    sphere   are    inscribed    in    a 
cylinder  of  revolution.     Compare  the  volumes  of  the  three  solids. 

16.  If   the  area  of  the   convex   surface   of   a   right   circular 
cone  is  twice  the  area  of  its  base,  prove  that  the  slant  height  of 
the  cone  is  equal  to  the  diameter  of  the  base. 

17.  If  the  specific  gravity  of  an  iron  ball  10"  in  diameter   is 
8.1,  what  is  its  weight? 

18.  An  iron  kettle  in  the  shape  of  a  hemisphere  has  a  <li;uu 
eter  of  3£  ft.      How  many  gallons  does  it  hold  when  full '     Wlion 
filled  to  a  depth  of  10  in.f 


EXERCISES  371 

19.  A  cube  and  a  sphere  have  equal  areas.     Which  has  the 
greater  volume? 

20.  Two  balls  are  of  the  same  material.     One,  weighing  12 
Ibs.,  has  a  diameter  equal  to   -J    that  of  the  other.     What  is  the 
weight  of  the  second  ball? 

21.  The  radius  of  the  base  of  a  right  circular  cone  is  5"     its 
volume  is  31    cu.  in.  and  the  number  of  square  inches  in  the  area 
of  its  convex  surface  is  equal  to  the  number  of  cubic  inches  in 
the  volume  of  the  cone.     What  is  its  altitude  and  its  slant  height  ? 

22.  Determine  the  locus  of  the  center  of  a  sphere  with  given 
radius,  such  that 

1.  its  surface  passes  through  a  given  point. 

2.  it  is  tangent  to  a  given  plane. 

3.  it  is  tangent  to  a  given  sect. 

4.  it  is  tangent  to  a  given  sphere. 

23.  Determine  the  locus  of  a  point  which  is 

1.  at  a  given  distance  from  a  fixed  point. 

2.  at  given  distances  from  two  fixed  points. 

3.  equally  distant  from  two  parallel  planes  and  at  a  fixed  dis- 
tance  from   a   given  point. 

4.  equally    distant    from    two    points    and    at    a    given    distance 
from  a  third  point. 

24.  Make  up  exercises  involving  loci  which  are  the  intersec- 
tions of  other  loci. 

25.  Find   the   center   of  a   sphere  with   a   given    radius,   such 
that 

1.  its  surface  passes  through  three  given  points. 

2.  it  is  tangent  to  a  given  plane  and  its  surface  passes  through 
two  given  points. 


372  SOLID  GEOMETRY 

3.  it   is   tangent    to    two    given   planes    and   its    surface    paspe ;» 
through   a   given   point. 

4.  it  is  tangent  to  three  given  planes. 

5.  it  is  tangent  to  a   given  sphere,  to  a  given  plane,   and  its 
surface  passes  through  a  given  point. 

6.  it   is  tangent  to   a   given  line,  a  given  plane,   and   a   given 
sphere. 


Make  other  exercises  in  this  group. 


26.  Find  the  pole  of  a  circle  determined 
by  three  points  on  the  surface  of  a  sphere. 

Sue.  It  can  be  done  in  a  manner 
similar  to  that  by  which  the  center  of 
a  circle  is  determined  by  three  points 
in  a  plane. 

27.  The    diameter   of   a   sphere    is   equal   to  the  altitude  of 
a   cone   of   revolution   and   of   a  cylinder   of   revolution,  the   radii 
of  the  three  solids  being  the  same.     Prove  that  the  volumes  of  the 
cone,  sphere,  and  cylinder  are  proportional  to  1,  2,  3  respectively. 

28.  An   orange,     3£    in.  in   diameter,   pulp   measure,  sells   at 
50  cents  a  dozen.     One  3  in.  in  diameter  sells  at  40  cents.     Which 
is  the  better  one  to  buy  if  the  quality  is  the  same?     If  the  larger 
orange  is  worth  50  cents,  what  is  the  smaller  one  worth? 

29.  If  a  3  in.  orange  sells  for  36  cents,  what  is  the  real  value 
of  a  4  in.  orange  of  same  quality? 

30.  What  part  of  a  spherical  surface  is  a  spherical  triangle 
each  angle  of  which  is  90°?     How  many  spherical  degrees  in  such 
a   triangle? 

31.  The  radius  of  a  sphere  is  10'  and  the  angles  of  a  spherical 
triangle  are  95°,  117°,  and  92°.     What  is  the  area  of  the  triangle? 

32.  What   is  the  area   of  the  earth's   surface,  assuming   the 
earth  to  be  a  sphere  with  a  radius  of  7,912  miles? 

33.  A  sphere  can  be  inscribed  in  a  cube. 


EXERCISES  373 

34.  By   planes    parallel    to    the    base    divide    a    pyramid    into 
four  equal  parts. 

35.  What  is  the  locus  of  the   vertex  of  a  pyramid  having  a 
fixed  volume  and  a  fixed  base  I 


REVIEW. 

791.     State  the    formula  for 

1.  The  volume  of  a  cone  of  revolution. 

2.  The  volume  of  a  cylinder  of  revolution. 

3.  The  volume  of  a  frustum  of  a  cone  of  revolution. 

4.  The  lateral  area  of  the  above  three  solids. 

5.  The  area  of  a  sphere. 

6.  The  area  of  a  spherical  triangle. 

7.  The  area  of  a  spherical  polygon  of  n  sides. 

8.  The  volume  of  a  sphere. 

9.  The  area  of  a  zone. 

10.  The  volume  of  a  spherical  sector. 

11.  The  volume  of  a  spherical  segment. 

12.  The  volume  of  a  spherical  pyramid. 

13.  The  volume  of  an  ungula. 


INDEX 


375 


Altitude     A 
Alteinate  interior  L 
Alternation 
Angle 


•    46 
30 
103 
4 
Angle    line    to    phme  ........  250 

Antecedents     ..............  101 

Apothem     .................  193 

Axiom    ....................    11 

Axis    sphere    ...............  332 

Base    A  ...................   20 

Base    G  ...................    ">•'* 

Bisector    ...................    11 

Broken  Line    ...............     4 

Circles     ...................   73 

Circle   of    a    sphere  .........  332 

Circumscribed   polygons  .....   87 

Composition     ...............  103 

Concurrent    lines  ...........    60 

Cone     .....  ................  310 

Congruent    ................   20 

Cosine    ....................  122 

Consequents     ..............  101 

Constant     .................  214 

Continuity    ................  134 

Converse     .................   30 

Cube    ......................  27(5 

Curved   line    ...............     4 

Cylinder     ..................  300 

Cylindrical     surface  .........  300 

Degree    ....................      7 

Depression   Z  of  ............  123 

Diagonal    ..................    53 

Diagonal   scale    ............  115 

Diameter    ................   53 

Dihedral  Z  .................  248 

Distance     ..................    4(i 

Division    ..................  104 

Division    external  .......  107.  146 

Division     harmonic..  ..100 


Exterior  Z 41 

Extremes     101 

Extreme  and  mean  ratio...  145 

External    tangent 159 

Foot    of    a    line 227 

Fourth    proportional 105 

Frustum,    cone 320 

Frustum,    pyramid 290 

Geometric    figure 2 

Geometric    solid 1 

Harmonic    division 109 

Homologous     110 

Horizontal   Z 123 

Incommensurable*    99,  213 

Indirect   proof 34 

Inscribed   polygons 87 

Inscribed   prisms 295 

Inscribed    sphere 342 

Interior    angles 36 

Internal    division 107 

Internal  tangent 159 

Inversion     103 

Isosceles    A 19 

Limit   of   a    variable 215 

Line   segment 3 

Locus    65 

Locus    (in  space) 234 

Magnitude    2,97 

Measurement    97 

Means     101 

Mean   proportion 140 

Median    47 

Numerical    measure 97 

Oblique   lines 11 

Opposite   interior   Z 41 

Obtuse    A 19 

Parallel    lines 34 

Parallel    pianos 240 

Parallelogram    52 


376 


INDEX 


Parallelepiped    270 

Parallels  postulate 34 

Perigon    6 

Perpendicular     bisector 25 

Perpendicular    lines 11 

Physical  solid.  . 1 

Plane    4 

Plane    figure 4 

Point    1 

Poles   333 

Polar   distance 334 

Polar    A 347 

Polygons     62 

Polyhedrals     260 

Polyhedrons     270 

Postulate     11 

Prism 271 

Projection   on   a   line 140 

Projection  on  a  plane 238 

Proportion    99 

Proportionals    140 

Pyramids     200 

Quadrilaterals    52 

Quantity    97 

Ratio    98 

Ray    3 

Regular    polygons 191 

Regular    prisms 272 

Regular    pyramids 290 

Right    angle 7 

Secant    line 36,74 

Sect   3 

Sector    .  87 


Segment    87 

Similar    polygons 110 

Similar  polyhedrons 302 

Similitude  of  polygons Ill 

Sine   121 

Slant  height    (pyramid)  ...  .290 

Slant    height    (cone) 320 

Sphere    330 

Spherical    angle 341 

Spherical    excess 355 

Spherical    polygons 344 

Spherical    sectors 362 

Spherical    segments 3(52 

Solid    geometry 223 

Straight   line 2 

Subtends    74 

Surface    I 

Symbols     13 

Symmetrical  polyhedrals. .  .  -260 
Symmetrical  spherical  poly- 
gons     345 

Tangent    (circle) 83 

Tangent    (cone) 320 

Tangent    (cylinder) 310 

Tangent  (sphere) 330 

Theorem 11 

Third   proportional 140 

Transversal    36 

Triangle    19 

Trigonometric  functions.  ..  .121 

Truncated    prism 275 

Truncated    pyramid 290 


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